Tüm alıştırma soruları

Soru 2241Soru →

In the standard $(x, y)$ coordinate plane, rectangle $ABCD$ has vertices $A(-3, 1)$ , $B(-1, 1)$ , $C(-1, 4)$ , and $D(-3, 4)$ . The rectangle is first reflected across the $y$ -axis and then translated $3$ units to the left and $2$ units down. What are the coordinates of the final image of vertex $C$ ?

(-2, 2)

(4, 2)

(-2, 6)

(-4, -6)

(-4, 2)

Cevabı ve açıklamayı göster

Cevap: $(-2, 2)$

Cevap

(-2, 2)

To find the final image of vertex

C(-1, 4)

, we first apply the reflection across the

y

-axis. The rule for reflecting a point across the

y

-axis is

(x, y) \rightarrow (-x, y)

. Applying this to

C(-1, 4)

gives

C'(1, 4)

. Next, we apply the translation of

3

units to the left and

2

units down. The rule for this translation is

(x, y) \rightarrow (x - 3, y - 2)

. Applying this to

C'(1, 4)

gives

C''(1 - 3, 4 - 2) = (-2, 2)

.

Adım Adım Çözüm

1

Identify the initial coordinates of vertex

C

.

C(-1, 4)

We need to find the final image of vertex

C

, so we start with its given coordinates.

2

Apply the reflection across the

y

-axis to vertex

C

.

C'(1, 4)

Reflecting a point

(x, y)

across the

y

-axis negates the x-coordinate, changing

(x, y)

to

(-x, y)

.

3

Apply the translation of

3

units to the left and

2

units down to the reflected point

C'

.

C''(-2, 2)

Translating a point

(x, y)

by

3

units to the left subtracts

3

from the x-coordinate (

x - 3

), and translating

2

units down subtracts

2

from the y-coordinate (

y - 2

).

Anahtar Kavram

Composite transformations in the coordinate plane involve applying multiple geometric transformations in a specific sequence.

Soru 2242Soru →

In the standard $(x, y)$ coordinate plane, line $L_1$ has a negative slope and passes through the point $(1, 4)$ . Line $L_2$ has a positive slope and passes through the point $(7, 3)$ . Both lines intersect the $x$ -axis at the same point $P$ . If the product of the slopes of $L_1$ and $L_2$ is $-\frac{4}{3}$ , and line $L_3$ is perpendicular to $L_1$ and passes through the point $(2, -2)$ , what is the $y$ -intercept of $L_3$ ?

Cevabı ve açıklamayı göster

Cevap: -3.5

Cevap

The

y

-intercept of line

L_3

is

-3.5

.

By writing the slope of the first line as

\frac{-4}{x_p - 1}

and the slope of the second line as

\frac{-3}{x_p - 7}

, their product is set to

-\frac{4}{3}

. Solving the resulting quadratic equation

(x_p - 4)^2 = 0

yields

x_p = 4

. Substituting this back gives a slope of

-\frac{4}{3}

for the first line. The perpendicular line must have a slope of

\frac{3}{4}

. Using the point-slope formula with point

(2, -2)

and slope

\frac{3}{4}

gives the line

y = \frac{3}{4}x - 3.5

, which crosses the

y

-axis at

-3.5

.

Adım Adım Çözüm

1

Express the slopes of

L_1

and

L_2

in terms of the unknown

x

-coordinate of point

P

.

m_1 = \frac{0 - 4}{x_p - 1} = \frac{-4}{x_p - 1}

and

m_2 = \frac{0 - 3}{x_p - 7} = \frac{-3}{x_p - 7}

Since

P

lies on the

x

-axis, its coordinates are

(x_p, 0)

. We apply the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

to the points on each line.

2

Set up the equation for the product of the slopes.

\left(\frac{-4}{x_p - 1}\right)\left(\frac{-3}{x_p - 7}\right) = -\frac{4}{3} \implies \frac{12}{(x_p - 1)(x_p - 7)} = -\frac{4}{3}

The problem states that the product of the slopes of

L_1

and

L_2

is

-\frac{4}{3}

.

3

Solve the equation for

x_p

.

x_p = 4

Cross-multiplying gives

36 = -4(x_p^2 - 8x_p + 7)

, which simplifies to

-9 = x_p^2 - 8x_p + 7 \implies x_p^2 - 8x_p + 16 = 0 \implies (x_p - 4)^2 = 0

.

4

Find the slope of

L_1

.

m_1 = \frac{-4}{4 - 1} = -\frac{4}{3}

Substituting

x_p = 4

back into the expression for

m_1

yields the slope of

L_1

.

5

Find the slope of

L_3

.

m_3 = \frac{3}{4}

Since

L_3

is perpendicular to

L_1

, its slope is the negative reciprocal of

m_1

.

6

Write the linear equation for

L_3

and determine its

y

-intercept.

y = \frac{3}{4}x - 3.5

, so the

y

-intercept is

-3.5

.

Using the point-slope form with point

(2, -2)

and slope

m_3 = \frac{3}{4}

, we get

y - (-2) = \frac{3}{4}(x - 2) \implies y + 2 = \frac{3}{4}x - 1.5 \implies y = \frac{3}{4}x - 3.5

.

Anahtar Kavram

The slope of a line is defined as

m = \frac{y_2 - y_1}{x_2 - x_1}

. Perpendicular lines have slopes that are negative reciprocals of each other, satisfying

m_1 \cdot m_2 = -1

. The

y

-intercept of a line is the value of

y

when

x = 0

.

Soru 2243Soru →

In the standard $(x, y)$ coordinate plane, a line passes through the points $(1, 3)$ and $(4, 9)$ . Is the statement that this line has a slope of $2$ true or false?

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

true

Applying the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

to the points

(1, 3)

and

(4, 9)

yields

\frac{9 - 3}{4 - 1} = \frac{6}{3} = 2

, which confirms that the statement is true.

Adım Adım Çözüm

1

Identify the coordinates of the two points on the line.

The points are

(x_1, y_1) = (1, 3)

and

(x_2, y_2) = (4, 9)

.

These coordinates are needed to apply the slope formula.

2

Apply the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

using the identified points.

m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2

.

This calculates the rate of change of the line, which is its slope.

3

Compare the calculated slope to the value in the statement.

The calculated slope is

2

, which matches the statement.

To determine whether the statement is true or false.

Anahtar Kavram

The slope of a line is the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line.

Tahmini Süre:45s

Soru 2244Soru →

In a coordinate plane, two perpendicular lines, $L_1$ and $L_2$ , intersect at a point on the positive $y$ -axis. The line $L_1$ passes through the point $(-4, 3)$ , and the line $L_2$ has an $x$ -intercept at $(2.5, 0)$ . What is the $y$ -coordinate of the intersection point of $L_1$ and $L_2$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The y-coordinate of the intersection point is 5.

The intersection point on the positive

y

-axis has coordinates

(0, 5)

. The slope of the line passing through

(-4, 3)

and

(0, 5)

is

0.5

. The slope of the line passing through

(2.5, 0)

and

(0, 5)

is

-2

. Since the product of these slopes is

0.5 \times (-2) = -1

, the lines are perpendicular.

Adım Adım Çözüm

1

Identify the coordinates of the intersection point.

The intersection point of

L_1

and

L_2

is

(0, b)

, where

b > 0

.

Since the intersection point lies on the positive

y

-axis, its

x

-coordinate is

0

and its

y

-coordinate

b

must be positive.

2

Calculate the slope of

L_1

.

The slope of

L_1

is

m_1 = \frac{b - 3}{0 - (-4)} = \frac{b - 3}{4}

.

The slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

is applied to the points

(-4, 3)

and

(0, b)

.

3

Calculate the slope of

L_2

.

The slope of

L_2

is

m_2 = \frac{b - 0}{0 - 2.5} = -\frac{b}{2.5}

.

The line

L_2

passes through the

y

-intercept

(0, b)

and its

x

-intercept

(2.5, 0)

.

4

Apply the perpendicular slope condition.

m_1 \cdot m_2 = -1 \implies \left(\frac{b - 3}{4}\right)\left(-\frac{b}{2.5}\right) = -1 \implies b(b - 3) = 10

.

Perpendicular lines have slopes that are negative reciprocals of each other, meaning their product is

-1

.

5

Solve the quadratic equation for

b

.

Expanding the equation gives

b^2 - 3b - 10 = 0

, which factors into

(b - 5)(b + 2) = 0

. Since

b > 0

, the only valid solution is

b = 5

.

Solving the factored equation yields

b = 5

or

b = -2

. The constraint that the intersection lies on the positive

y

-axis rules out the negative value.

Anahtar Kavram

Perpendicular line slopes and coordinate intercepts

Tahmini Süre:2m 30s

Soru 2245Soru →

For all real numbers $x$ and $y$ such that $x > y > 0$ , which of the following expressions is equivalent to $\log_5(x^2 - y^2) - \log_5(x - y)$ ?

\log_5(x-y)

\log_5(x^2 - y^2 - x + y)

\log_5(x) - \log_5(y)

\log_5(x+y)

x + y

Cevabı ve açıklamayı göster

Cevap: $\log_5(x+y)$

Cevap

The expression is equivalent to \log_5(x + y)

The correct answer is found by using the quotient property of logarithms to combine the terms:

\log_5(x^2 - y^2) - \log_5(x - y) = \log_5\left(\frac{x^2 - y^2}{x - y}\right)

. Factoring the difference of squares in the numerator yields

\log_5\left(\frac{(x-y)(x+y)}{x-y}\right)

. Canceling the common factor

(x-y)

simplifies the expression to

\log_5(x+y)

.

Adım Adım Çözüm

1

Combine the logarithms using the quotient property of logarithms:

\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)

.

\log_5\left(\frac{x^2 - y^2}{x - y}\right)

This simplifies the difference between two logarithms with the same base into a single logarithm.

2

Factor the difference of squares in the numerator:

x^2 - y^2 = (x - y)(x + y)

.

\log_5\left(\frac{(x - y)(x + y)}{x - y}\right)

This allows for the cancellation of common factors in the fraction.

3

Cancel the common factor

(x - y)

from both the numerator and the denominator.

\log_5(x + y)

Since

x > y > 0

, the term

x - y

is non-zero, making the division valid.

Anahtar Kavram

Quotient property of logarithms and difference of squares factoring

Alternatif Yöntem

Instead of applying the quotient property first, one can factor the argument of the first term using the difference of squares:

\log_5(x^2 - y^2) = \log_5((x - y)(x + y))

. Next, apply the product property of logarithms to split this term:

\log_5(x - y) + \log_5(x + y)

. Substituting this back into the original expression gives

\log_5(x - y) + \log_5(x + y) - \log_5(x - y)

, which simplifies directly to

\log_5(x + y)

.

Tahmini Süre:1m 0s

Soru 2246Soru →

In the standard $(x, y)$ coordinate plane, the midpoint of the line segment with endpoints $A(2, a)$ and $B(b, -3)$ lies on the line $y = 3x - 4$ . If the distance between $A$ and $B$ is $5\sqrt{2}$ and $a > 0$ , what is the value of $a + b$ ?

13

-11

5

3

1

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of

a + b

is

5

.

The midpoint of the segment with endpoints

(2, a)

and

(b, -3)

is

\left(\frac{2+b}{2}, \frac{a-3}{2}\right)

. Substituting these coordinates into the line

y = 3x - 4

gives

\frac{a-3}{2} = 3\left(\frac{2+b}{2}\right) - 4

, which simplifies to

a = 3b + 1

. The squared distance between the endpoints is

(b-2)^2 + (-3-a)^2 = (5\sqrt{2})^2 = 50

. Substituting

a = 3b + 1

into the distance equation yields

(b-2)^2 + (3b+4)^2 = 50

, which simplifies to the quadratic equation

b^2 + 2b - 3 = 0

. Factoring gives

(b+3)(b-1) = 0

, so

b = 1

or

b = -3

. The constraint

a > 0

means

a = 3(1) + 1 = 4

is the only valid solution. Therefore, the sum is

4 + 1 = 5

.

Adım Adım Çözüm

1

Find the midpoint of the segment

AB

using the midpoint formula.

M = \left(\frac{2+b}{2}, \frac{a-3}{2}\right)

The midpoint is defined as the average of the x-coordinates and the average of the y-coordinates.

2

Substitute the midpoint coordinates into the equation of the line

y = 3x - 4

to relate

a

and

b

.

a = 3b + 1

Since the midpoint lies on the line, its coordinates must satisfy the line's equation.

3

Use the distance formula to set up an equation for the distance between

A

and

B

.

(b-2)^2 + (a+3)^2 = 50

The squared distance between

(2, a)

and

(b, -3)

is

(5\sqrt{2})^2 = 50

.

4

Substitute the relation

a = 3b + 1

into the distance equation and solve for

b

.

b = 1

or

b = -3

This yields the quadratic equation

b^2 + 2b - 3 = 0

, which factors as

(b+3)(b-1) = 0

.

5

Apply the constraint

a > 0

to find the correct values of

a

and

b

.

a = 4

and

b = 1

If

b = -3

, then

a = 3(-3) + 1 = -8

, which is not greater than

0

. If

b = 1

, then

a = 3(1) + 1 = 4 > 0

.

6

Calculate the sum

a + b

.

4 + 1 = 5

To find the final requested value.

Anahtar Kavram

Using the distance and midpoint formulas in combination with linear equations to solve for unknown coordinates.

Tahmini Süre:2m 30s

Soru 2247Soru →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 + y^2 - 4x + 6y - 12 = 0$ . What is the radius of this circle?

25

2\sqrt{3}

5

8

\sqrt{17}

Cevabı ve açıklamayı göster

Cevap: $5$

Cevap

The radius of the circle is 5.

The correct answer is 5. Grouping the

x

-terms and

y

-terms of the equation

x^2 + y^2 - 4x + 6y - 12 = 0

gives

(x^2 - 4x) + (y^2 + 6y) = 12

. Completing the square requires adding

(\frac{-4}{2})^2 = 4

and

(\frac{6}{2})^2 = 9

to both sides, yielding

(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9

, which simplifies to

(x-2)^2 + (y+3)^2 = 25

. In the standard circle equation form

(x-h)^2 + (y-k)^2 = r^2

,

r^2 = 25

, meaning the radius

r

is

\sqrt{25} = 5

.

Adım Adım Çözüm

1

Group the

x

-terms and

y

-terms, and move the constant term to the right side of the equation.

(x^2 - 4x) + (y^2 + 6y) = 12

This prepares the quadratic expression to be written in standard circle form by completing the square.

2

Find the constant values needed to complete the square for both the

x

and

y

variables. Add these constants to both sides of the equation.

(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9

The constant for

x

is

(\frac{-4}{2})^2 = 4

, and the constant for

y

is

(\frac{6}{2})^2 = 9

. Adding them to both sides maintains equality.

3

Factor the perfect square trinomials on the left side and simplify the right side.

(x-2)^2 + (y+3)^2 = 25

This puts the equation in the standard form

(x-h)^2 + (y-k)^2 = r^2

.

4

Identify

r^2

from the standard form and solve for the radius

r

.

r = \sqrt{25} = 5

Taking the square root of the constant on the right side yields the radius of the circle.

Anahtar Kavram

Rewriting a circle's equation from general form to standard form by completing the square to find its radius.

Alternatif Yöntem

For a circle given in the general form

x^2 + y^2 + Dx + Ey + F = 0

, the radius can be calculated directly using the formula

r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}

. Substituting the coefficients

D = -4

,

E = 6

, and

F = -12

gives:

r = \frac{1}{2}\sqrt{(-4)^2 + 6^2 - 4(-12)} = \frac{1}{2}\sqrt{16 + 36 + 48} = \frac{1}{2}\sqrt{100} = \frac{1}{2}(10) = 5

.

Tahmini Süre:1m 30s

Soru 2248Soru →

In $\triangle ABC$ , the measure of interior angle $\angle A$ is $45^\circ$ , and the exterior angle at vertex $B$ measures $125^\circ$ . What is the measure, in degrees, of interior angle $\angle C$ ?

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The measure of interior angle

\angle C

is

80

degrees.

According to the Exterior Angle Theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles. In this case, the exterior angle at vertex

B

measures

125^\circ

, and one of its remote interior angles,

\angle A

, measures

45^\circ

. The other remote interior angle is

\angle C

. Setting up the equation:

125^\circ = 45^\circ + \text{m}\angle C

. Solving for the measure of

\angle C

gives

125^\circ - 45^\circ = 80^\circ

.

Adım Adım Çözüm

1

Use the Exterior Angle Theorem to relate the given angles.

The exterior angle at vertex

B

(

125^\circ

) is equal to the sum of the remote interior angles,

\angle A

and

\angle C

. This gives the equation:

125^\circ = 45^\circ + \text{m}\angle C

.

The Exterior Angle Theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.

2

Solve the equation for the measure of

\angle C

.

\text{m}\angle C = 125^\circ - 45^\circ = 80^\circ

Subtract

45^\circ

from both sides of the equation to isolate the measure of

\angle C

.

Anahtar Kavram

Exterior Angle Theorem

Soru 2249Soru →

In the standard $(x, y)$ coordinate plane, line $L_1$ passes through the points $(-1, 2)$ and $(3, 14)$ . If line $L_2$ is parallel to line $L_1$ , what is the slope of line $L_2$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The slope of line

L_2

is

3

.

To find the slope of line

L_2

, we calculate the slope of line

L_1

using the formula

m = \frac{y_2 - y_1}{x_2 - x_1}

. Substituting the coordinates

(-1, 2)

and

(3, 14)

yields

m = \frac{14 - 2}{3 - (-1)} = \frac{12}{4} = 3

. Because parallel lines have the same slope, the slope of line

L_2

is also

3

.

Adım Adım Çözüm

1

Calculate the slope of line

L_1

using the coordinates of the two given points,

(-1, 2)

and

(3, 14)

.

m = 3

The slope formula is

m = \frac{y_2 - y_1}{x_2 - x_1}

, which evaluates to

m = \frac{14 - 2}{3 - (-1)} = \frac{12}{4} = 3

.

2

Determine the slope of line

L_2

based on its relationship to line

L_1

.

The slope of line

L_2

is

3

.

Parallel lines always have identical slopes.

Anahtar Kavram

Parallel lines have the same slope.

Soru 2250Soru →

For all positive real numbers $x$ and $y$ , consider the algebraic expression:

\frac{\left((x + y)y^2\right)^{1/2} (x + y)^{2/3}}{(x + y)^{-1/2}y}

Which of the following is equivalent to this expression?

(x + y)^{5/6}

x^{5/3} + y^{5/3}

(x + y)^{5/3}

(x + y)^{11/10}

(x + y)^{5/3} y

Cevabı ve açıklamayı göster

Cevap: $(x + y)^{5/3}$

Cevap

(x + y)^{5/3}

The correct answer is obtained by applying the properties of exponents step-by-step. First, apply the power of a product rule to the term

((x+y)y^2)^{1/2}

to get

(x+y)^{1/2}y

. Next, combine the terms in the numerator using the product rule of exponents to get

(x+y)^{7/6}y

. Finally, divide by the denominator using the quotient rule of exponents, which cancels the

y

terms and results in

(x+y)^{5/3}

.

Adım Adım Çözüm

1

Apply the power of a product property to simplify the first term in the numerator.

\left((x + y)y^2\right)^{1/2} = (x + y)^{1/2}(y^2)^{1/2} = (x + y)^{1/2}y

To distribute the exponent

1/2

to each factor inside the parentheses.

2

Multiply the simplified term by the other factor in the numerator.

(x + y)^{1/2}y \cdot (x + y)^{2/3} = (x + y)^{1/2 + 2/3}y = (x + y)^{7/6}y

To combine bases of the same value by adding their exponents:

\frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6}

.

3

Divide the simplified numerator by the denominator.

\frac{(x + y)^{7/6}y}{(x + y)^{-1/2}y} = (x + y)^{7/6 - (-1/2)} = (x + y)^{7/6 + 3/6} = (x + y)^{10/6} = (x + y)^{5/3}

To subtract the exponent in the denominator from the exponent in the numerator and cancel the common factor

y

.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions

Alternatif Yöntem

Instead of simplifying the numerator first, we can divide the terms inside the parentheses first if we rewrite the expression, but the standard method of simplifying the numerator and then performing division is the most direct.

Tahmini Süre:1m 30s

Soru 2251Soru →

If $\log_2(x + 5) + \log_2(x - 1) = 4$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The correct answer is 3.

The correct answer is 3. Combining the logarithmic terms using the product property gives

\log_2((x + 5)(x - 1)) = 4

. Converting this to exponential form yields

(x + 5)(x - 1) = 2^4 = 16

. Expanding and rewriting in standard form gives

x^2 + 4x - 21 = 0

, which factors as

(x + 7)(x - 3) = 0

. This gives potential solutions of

x = 3

and

x = -7

. However,

x = -7

results in negative arguments for the logarithms in the original equation, making it extraneous. Thus, the only valid solution is 3.

Adım Adım Çözüm

1

Apply the product property of logarithms to combine the terms on the left side of the equation.

\log_2((x + 5)(x - 1)) = 4

The sum of logarithms with the same base is equal to the logarithm of their product:

\log_b(A) + \log_b(B) = \log_b(AB)

.

2

Convert the logarithmic equation to its equivalent exponential form.

(x + 5)(x - 1) = 2^4

By definition,

\log_b(Y) = C

is equivalent to

b^C = Y

.

3

Expand the product and simplify the equation into standard quadratic form.

x^2 + 4x - 21 = 0

Expanding

(x + 5)(x - 1)

yields

x^2 + 4x - 5

. Subtracting 16 from both sides gives the standard quadratic form

Ax^2 + Bx + C = 0

.

4

Factor the quadratic equation to find the potential values of

x

.

(x + 7)(x - 3) = 0

, so

x = -7

or

x = 3

We need two numbers that multiply to

-21

and add to

4

, which are

7

and

-3

.

5

Check the potential solutions against the domain of the original logarithmic equation.

x = 3

The arguments of the logarithms,

x + 5

and

x - 1

, must be strictly positive. For

x = -7

, the arguments are negative, so

x = -7

is an extraneous solution. For

x = 3

, both arguments are positive, so

x = 3

is the only valid solution.

Anahtar Kavram

Solving logarithmic equations by applying logarithmic properties and converting to exponential form, while checking for extraneous solutions.

Soru 2252Soru →

In the standard $(x, y)$ coordinate plane, a line segment has endpoints $A(-3, 2)$ and $B(1, 4)$ . The segment is reflected across the line $y = -x$ to form segment $A'B'$ , which is then translated $5$ units to the right and $3$ units down to form segment $A''B''$ . What are the coordinates of the midpoint of segment $A''B''$ ?

(2, -2)

(-8, 4)

(8, -4)

(0, -4)

(6, 0)

Cevabı ve açıklamayı göster

Cevap: $(2, -2)$

Cevap

(2, -2)

The correct answer is

(2, -2)

. The midpoint of the original segment

AB

is calculated as

M(-1, 3)

. Reflecting

M

across the line

y = -x

yields

M'(-3, 1)

. Translating

M'

by

5

units right and

3

units down results in the final midpoint

M''(2, -2)

.

Adım Adım Çözüm

1

Find the coordinates of the midpoint of the original segment

AB

.

The midpoint

M

of

AB

is calculated using the midpoint formula:

M = \left(\frac{-3 + 1}{2}, \frac{2 + 4}{2}\right) = (-1, 3)

.

Since translations, reflections, and dilations preserve midpoints, the midpoint of the final image segment

A''B''

is the same point obtained by applying the transformations to the midpoint of the original segment

AB

.

2

Apply the reflection across the line

y = -x

to the midpoint

M(-1, 3)

.

Using the reflection rule

(x, y) \rightarrow (-y, -x)

, the reflected midpoint is

M' = (-3, 1)

.

A reflection across

y = -x

swaps the coordinates of a point and negates both values.

3

Apply the translation of

5

units right and

3

units down to the point

M'(-3, 1)

.

Using the translation rule

(x, y) \rightarrow (x + 5, y - 3)

, the final midpoint is

M'' = (-3 + 5, 1 - 3) = (2, -2)

.

Translating a point right increases the x-coordinate, and translating it down decreases the y-coordinate.

Anahtar Kavram

Applying composite transformations (reflection across

y = -x

followed by translation) to geometric objects and midpoints in the coordinate plane.

Tahmini Süre:2m 0s

Soru 2253Soru →

In acute triangle $ABC$ , the measure of $\angle A$ is $74^\circ$ . The altitude from vertex $B$ to side $AC$ and the altitude from vertex $C$ to side $AB$ intersect at point $H$ inside the triangle. What is the measure, in degrees, of $\angle BHC$ ?

Cevabı ve açıklamayı göster

Cevap: 106

Cevap

The measure of

\angle BHC

is 106 degrees.

In quadrilateral

AEHD

, the sum of the angles is

360^\circ

. Since the angles at

E

and

D

are

90^\circ

because they are formed by altitudes, the sum of

\angle A

and

\angle EHD

must be

180^\circ

. Thus,

\angle EHD = 180^\circ - 74^\circ = 106^\circ

. Since

\angle BHC

and

\angle EHD

are vertical angles,

\angle BHC = 106^\circ

.

Adım Adım Çözüm

1

Define the intersection points of the altitudes with the opposite sides.

Let the altitude from vertex

B

intersect side

AC

at point

D

, and let the altitude from vertex

C

intersect side

AB

at point

E

. Therefore,

\angle ADH = 90^\circ

and

\angle AEH = 90^\circ

.

Altitudes by definition are perpendicular to the sides they intersect, forming

90^\circ

angles.

2

Analyze the sum of interior angles in the quadrilateral

AEHD

.

The sum of the interior angles of a quadrilateral is

360^\circ

, so

\angle A + \angle AEH + \angle EHD + \angle ADH = 360^\circ

.

Any quadrilateral can be split into two triangles, making the sum of its interior angles

360^\circ

.

3

Substitute the known angle measures to find the measure of

\angle EHD

.

74^\circ + 90^\circ + \angle EHD + 90^\circ = 360^\circ \implies 254^\circ + \angle EHD = 360^\circ \implies \angle EHD = 106^\circ

.

Solving the linear equation for the unknown angle

\angle EHD

.

4

Relate

\angle EHD

to the target angle

\angle BHC

.

Since line segments

BD

and

CE

intersect at

H

, the angles

\angle BHC

and

\angle EHD

are vertical angles, so

\angle BHC = \angle EHD = 106^\circ

.

Vertical angles are equal in measure.

Anahtar Kavram

The sum of angles in quadrilaterals, the definition of altitudes, and vertical angles within triangles.

Tahmini Süre:2m 0s

Soru 2254Soru →

In the standard $(x, y)$ coordinate plane, the line described by the equation $2x - 3y = 10$ is translated $2$ units to the left and $1$ unit up. What is the $y$ -intercept of the translated line?

-\frac{17}{3}

-3

-\frac{7}{3}

-\frac{5}{3}

-1

Cevabı ve açıklamayı göster

Cevap: $-1$

Cevap

The correct answer is

-1

.

The correct answer is

-1

. Translating a line 2 units to the left shifts every point on the line by

-2

in the

x

-direction, which is represented algebraically by substituting

x + 2

for

x

. Translating it 1 unit up shifts every point by

+1

in the

y

-direction, represented by substituting

y - 1

for

y

. Substituting these into the original equation

2x - 3y = 10

yields

2(x + 2) - 3(y - 1) = 10

. Simplifying this equation gives

2x + 4 - 3y + 3 = 10

, which becomes

2x - 3y = 3

. To find the

y

-intercept of this new line, we set

x = 0

, giving

-3y = 3

, which solves to

y = -1

. Alternatively, we can find the original

y

-intercept of

(0, -10/3)

. Under the translation, this point shifts to

(0 - 2, -10/3 + 1) = (-2, -7/3)

. Since the slope of the line remains

2/3

, the new line equation is

y - (-7/3) = (2/3)(x - (-2))

, which simplifies to

y = (2/3)x - 1

, confirming the

y

-intercept is

-1

.

Adım Adım Çözüm

1

Express the translation of the coordinates mathematically. Translating a line 2 units to the left replaces

x

with

x + 2

. Translating it 1 unit up replaces

y

with

y - 1

.

The substitution expressions are

x \to x + 2

and

y \to y - 1

.

This sets up the algebraic transformation of the original line's equation.

2

Substitute the expressions into the original equation

2x - 3y = 10

and simplify.

2(x + 2) - 3(y - 1) = 10 \implies 2x + 4 - 3y + 3 = 10 \implies 2x - 3y + 7 = 10 \implies 2x - 3y = 3

.

This derives the equation of the translated line.

3

Find the

y

-intercept of the translated line

2x - 3y = 3

by setting

x = 0

.

2(0) - 3y = 3 \implies -3y = 3 \implies y = -1

.

The

y

-intercept is the point on the line where

x = 0

.

Anahtar Kavram

Translating linear equations in the standard coordinate plane

Tahmini Süre:1m 30s

Soru 2255Soru →

A circle in the standard $(x,y)$ coordinate plane is defined by the equation $3x^2 + 3y^2 - 24x + 12y + k = 0$ , where $k$ is a constant. If the circle is tangent to the line $y = 2$ , what is the value of $k$ ?

-48

0

12

48

60

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12

The correct answer is 12. Dividing the given equation

3x^2 + 3y^2 - 24x + 12y + k = 0

by

3

gives

x^2 + y^2 - 8x + 4y + \frac{k}{3} = 0

. Completing the square for both variables yields

(x-4)^2 + (y+2)^2 = 20 - \frac{k}{3}

. This represents a circle centered at

(4, -2)

with a radius squared of

r^2 = 20 - \frac{k}{3}

. A circle centered at

(4, -2)

that is tangent to the line

y = 2

has a radius equal to the vertical distance between

y = -2

and

y = 2

, which is

|-2 - 2| = 4

. Thus,

r^2 = 16

. Setting

20 - \frac{k}{3} = 16

yields

\frac{k}{3} = 4

, which gives

k = 12

.

Adım Adım Çözüm

1

Divide the entire equation by 3 to normalize the coefficients of the quadratic terms.

x^2 + y^2 - 8x + 4y + \frac{k}{3} = 0

The standard equation of a circle requires the coefficients of

x^2

and

y^2

to be 1.

2

Complete the square for the

x

and

y

terms.

(x - 4)^2 - 16 + (y + 2)^2 - 4 + \frac{k}{3} = 0 \Rightarrow (x - 4)^2 + (y + 2)^2 = 20 - \frac{k}{3}

This rewrites the equation in the standard form

(x - h)^2 + (y - k)^2 = r^2

to identify the center and radius.

3

Identify the center of the circle and set up the radius relationship.

Center is

(4, -2)

and the radius squared is

r^2 = 20 - \frac{k}{3}

.

Comparing

(x - 4)^2 + (y + 2)^2 = 20 - \frac{k}{3}

to the standard form gives the center

(h, k) = (4, -2)

and

r^2

.

4

Determine the radius using the tangency condition with the line

y = 2

.

Radius

r = |y_{\text{center}} - 2| = |-2 - 2| = 4

Since the circle is tangent to the horizontal line

y = 2

, the radius is the vertical distance from the center's y-coordinate to the line.

5

Solve for the constant

k

using the radius squared value.

r^2 = 16 \Rightarrow 20 - \frac{k}{3} = 16 \Rightarrow 4 = \frac{k}{3} \Rightarrow k = 12

Equating the two expressions for

r^2

allows us to solve for

k

.

Anahtar Kavram

Converting the general equation of a circle to standard form by completing the square and using geometric tangency conditions to solve for unknown constants.

Tahmini Süre:2m 0s

Soru 2256Soru →

In the standard $(x, y)$ coordinate plane, line segment $CD$ has endpoints $C(-2, 2)$ and $D(-2, 8)$ . If segment $CD$ is reflected across the $y$ -axis to form segment $C'D'$ , what are the coordinates of the midpoint of segment $C'D'$ ?

(2, 5)

(-2, 5)

(-2, -5)

(2, -5)

(5, 2)

Cevabı ve açıklamayı göster

Cevap: $(2, 5)$

Cevap

The coordinate point

(2, 5)

The correct answer is

(2, 5)

because the midpoint of the original segment

CD

is located at

(-2, 5)

. When a point is reflected across the

y

-axis, its

x

-coordinate changes sign while its

y

-coordinate remains the same. Therefore, the image of the midpoint is

(2, 5)

.

Adım Adım Çözüm

1

Find the midpoint of the original segment

CD

.

Midpoint

M = \left(\frac{-2 + (-2)}{2}, \frac{2 + 8}{2}\right) = (-2, 5)

The midpoint formula is

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

.

2

Apply the reflection across the

y

-axis to the midpoint.

M' = (-(-2), 5) = (2, 5)

Reflecting a point

(x, y)

across the

y

-axis maps it to

(-x, y)

.

Anahtar Kavram

Reflection of points and segments across the axes in the coordinate plane

Alternatif Yöntem

Instead of finding the midpoint first, you can reflect the endpoints

C(-2, 2)

and

D(-2, 8)

across the

y

-axis to get

C'(2, 2)

and

D'(2, 8)

. Then, apply the midpoint formula to these new coordinates to get the midpoint:

\left(\frac{2 + 2}{2}, \frac{2 + 8}{2}\right) = (2, 5)

.

Tahmini Süre:45s

Soru 2257Soru →

A water tank is being drained at a constant rate. After $2$ hours of draining, the tank contains $140$ gallons of water, and after $5$ hours, it contains $80$ gallons. When the relationship between the time spent draining, $t$ , and the volume of water, $v$ , is plotted on a standard coordinate plane with $t$ on the horizontal axis, what is the slope of the resulting line?

-60

-20

-\frac{1}{20}

20

60

Cevabı ve açıklamayı göster

Cevap: $-20$

Cevap

The slope of the line is

-20

.

The correct answer is

-20

. The slope represents the rate of change of the volume of water with respect to time. Using the points

(2, 140)

and

(5, 80)

, the slope is calculated as the change in volume divided by the change in time:

\frac{80 - 140}{5 - 2} = \frac{-60}{3} = -20

.

Adım Adım Çözüm

1

Identify the coordinates of the two data points from the problem context.

The two points are

(2, 140)

and

(5, 80)

, where time

t

represents the horizontal coordinate and volume

v

represents the vertical coordinate.

The problem defines the horizontal axis as representing time

t

and the vertical axis as representing volume

v

.

2

Apply the slope formula

m = \frac{v_2 - v_1}{t_2 - t_1}

using the coordinates from Step 1.

m = \frac{80 - 140}{5 - 2} = \frac{-60}{3} = -20

.

Calculating the ratio of change in the vertical axis to the change in the horizontal axis gives the slope of the line.

Anahtar Kavram

Slope as a constant rate of change in a linear relationship

Tahmini Süre:1m 0s

Soru 2258Soru →

In the standard $(x, y)$ coordinate plane, line $L$ has a $y$ -intercept of $4$ . When line $L$ is translated horizontally to the right by $3$ units, the $y$ -intercept of the resulting line is $6$ units less than the $y$ -intercept of line $L$ . What is the $x$ -intercept of line $L$ ?

2

-8

-2

-\frac{1}{2}

-6

Cevabı ve açıklamayı göster

Cevap: $-2$

Cevap

The

x

-intercept of line

L

is

-2

.

The original line

L

can be written as

y = mx + 4

. Translating the line horizontally to the right by

3

units replaces

x

with

x - 3

, resulting in the equation

y = m(x - 3) + 4 = mx - 3m + 4

. The

y

-intercept of this new line is

4 - 3m

. We are given that this new

y

-intercept is

6

units less than the original

y

-intercept of

4

, which gives the equation

4 - 3m = 4 - 6

. Solving for

m

yields

-3m = -6 \implies m = 2

. With a slope of

2

and a

y

-intercept of

4

, the equation of the original line

L

is

y = 2x + 4

. To find the

x

-intercept, we set

y = 0

, which gives

2x + 4 = 0 \implies x = -2

.

Adım Adım Çözüm

1

Write the general equation of line

L

in slope-intercept form.

y = mx + 4

, where

m

is the slope of line

L

.

The

y

-intercept of line

L

is given as

4

, which represents the constant term

b

in

y = mx + b

.

2

Apply the horizontal translation of

3

units to the right.

y = m(x - 3) + 4

, which simplifies to

y = mx - 3m + 4

.

Translating a function horizontally to the right by

h

units replaces

x

with

x - h

.

3

Determine the slope

m

using the change in

y

-intercept.

The new

y

-intercept is

4 - 3m

. Since this is

6

units less than the original

y

-intercept of

4

, we set up the equation

4 - 3m = 4 - 6 \implies -3m = -6 \implies m = 2

.

We relate the algebraic expression for the new

y

-intercept to the given numerical relationship.

4

Find the

x

-intercept of line

L

by setting

y = 0

.

2x + 4 = 0 \implies 2x = -4 \implies x = -2

.

The

x

-intercept is the value of

x

when the line crosses the

x

-axis, meaning

y = 0

.

Anahtar Kavram

Linear Equations and Graphing

Soru 2259Soru →

A circle in the first quadrant of the standard $(x,y)$ coordinate plane is tangent to the $x$ -axis and is also tangent to the line $y = \frac{4}{3}x$ . If the center of the circle lies on the line with equation $y = 3x - 10$ , what is the radius of the circle?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The radius of the circle is 2.

The correct answer is obtained by determining that the center of a circle tangent to the

x

-axis in the first quadrant has the form

(h, R)

where

R

is the radius. Using the distance from this point to the line

4x - 3y = 0

gives

h = 2R

to ensure the center remains in the first quadrant. Substituting

(2R, R)

into the line

y = 3x - 10

yields

R = 3(2R) - 10

, which solves to

R = 2

.

Adım Adım Çözüm

1

Determine the relation between the circle's center coordinates and its radius.

The center is

(h, R)

where

R

is the radius.

Because the circle is tangent to the

x

-axis and lies in the first quadrant, the y-coordinate of its center must equal its radius.

2

Apply the point-to-line distance formula from the center to the line

y = \frac{4}{3}x

.

The relation is

|4h - 3R| = 5R

.

The distance from the center

(h, R)

to the line

4x - 3y = 0

must equal the radius

R

.

3

Solve the absolute value equation for

h

in terms of

R

.

Since

h > 0

, we find

h = 2R

.

The positive case

4h - 3R = 5R

gives

h = 2R

, whereas the negative case

4h - 3R = -5R

gives a negative

h

which violates the first-quadrant condition.

4

Substitute the center coordinates

(2R, R)

into the line

y = 3x - 10

and solve for

R

.

R = 2

.

The center lies on this line, so its coordinates must satisfy the line's equation.

Anahtar Kavram

Equations and Graphs of Circles

Soru 2260Soru →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $(x - 1)^2 + (y - 2)^2 = 10$ , and a line is defined by the equation $y = 3x - 1$ . The line intersects the circle at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the sum of the $y$ -coordinates of these two points of intersection?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The sum of the

y

-coordinates of the intersection points is 4.

Substituting

y = 3x - 1

into the circle's equation gives

(x-1)^2 + (3x-3)^2 = 10

. Factoring out 3 from the second term yields

(x-1)^2 + 9(x-1)^2 = 10

, which simplifies to

10(x-1)^2 = 10

, or

(x-1)^2 = 1

. Solving for

x

gives

x = 0

and

x = 2

. Substituting these values into the linear equation gives the corresponding

y

-coordinates:

y = -1

when

x = 0

, and

y = 5

when

x = 2

. The sum of these

y

-coordinates is

5 + (-1) = 4

.

Adım Adım Çözüm

1

Substitute the expression for

y

from the linear equation into the circle's equation.

(x-1)^2 + (3x-3)^2 = 10

To reduce the system of two equations to a single equation in terms of

x

.

2

Factor out 3 from the term

(3x-3)

and simplify the equation.

10(x-1)^2 = 10

, which simplifies to

(x-1)^2 = 1

To solve for the

x

-coordinates of the intersection points.

3

Solve the simplified quadratic equation for

x

.

x_1 = 2

and

x_2 = 0

To find the

x

-coordinates of the two intersection points.

4

Substitute the

x

-values back into the linear equation

y = 3x - 1

to find the corresponding

y

-coordinates.

y_1 = 3(2) - 1 = 5

and

y_2 = 3(0) - 1 = -1

To determine the

y

-coordinates of the intersection points

(2, 5)

and

(0, -1)

.

5

Add the two

y

-coordinates together.

5 + (-1) = 4

To find the sum of the

y

-coordinates as requested by the question.

Anahtar Kavram

Solving systems of linear and circular equations by substitution

Alternatif Yöntem

Instead of finding the individual coordinates, substitute

y_1 = 3x_1 - 1

and

y_2 = 3x_2 - 1

to write the sum as

y_1 + y_2 = 3(x_1 + x_2) - 2

. Expanding the substitution equation gives

10x^2 - 20x = 0

. By Vieta's formulas, the sum of the roots

x_1 + x_2 = -(-20)/10 = 2

. Substituting this back gives

y_1 + y_2 = 3(2) - 2 = 4

.

Tahmini Süre:1m 30s

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Alternatif Yöntem

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Alternatif Yöntem

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