Tüm alıştırma soruları

Match each chemical compound or process related to water hardness with its correct description or function.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Calcium hydrogencarbonate matches with causes temporary hardness; Calcium tetraoxosulfate(VI) matches with causes permanent hardness; Boiling matches with removes temporary hardness only; Addition of washing soda matches with removes both temporary and permanent hardness.

Calcium hydrogencarbonate is soluble and decomposes upon heating to form insoluble calcium carbonate, causing temporary hardness. Calcium tetraoxosulfate(VI) stays dissolved when boiled, causing permanent hardness. Boiling specifically precipitates temporary hardness salts. Washing soda contains carbonate ions which precipitate calcium ions from both hydrogencarbonates and sulfates, removing both temporary and permanent hardness.

Adım Adım Çözüm

1

Identify the cause of temporary hardness

Calcium hydrogencarbonate (

\text{Ca(HCO}_3\text{)}_2

) causes temporary hardness because it thermally decomposes when heated.

Temporary hardness is due to soluble hydrogen trioxocarbonates of calcium and magnesium.

2

Identify the cause of permanent hardness

Calcium tetraoxosulfate(VI) (

\text{CaSO}_4

) causes permanent hardness because boiling does not precipitate it.

Permanent hardness is due to soluble sulfates and chlorides of calcium and magnesium.

3

Determine the effect of boiling

Boiling removes temporary hardness by converting dissolved hydrogencarbonates into insoluble calcium carbonate precipitate.

Thermal decomposition:

\text{Ca(HCO}_3\text{)}_2\text{(aq)} \rightarrow \text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

.

4

Determine the effect of adding washing soda

Washing soda (

\text{Na}_2\text{CO}_3

) removes both temporary and permanent hardness.

Carbonate ions precipitate

\text{Ca}^{2+}

ions as

\text{CaCO}_3\text{(s)}

regardless of whether they originated from hydrogencarbonate or sulfate salts.

Anahtar Kavram

Causes, types, and chemical methods for removal of water hardness

Tahmini Süre:45s

Soru 3562Soru →

In the reaction between octasulfur, $S_8$ , and concentrated trioxonitrate(V) acid, sulfur is oxidized to tetraoxosulfate(VI) acid, $H_2SO_4$ . What are the oxidation numbers of sulfur in $S_8$ and $H_2SO_4$ respectively?

0

and

+6

-2

and

+6

+8

and

+6

0

and

+4

Cevabı ve açıklamayı göster

Cevap: $0$ and $+6$

Cevap

The oxidation numbers of sulfur in

S_8

and

H_2SO_4

are

0

and

+6

respectively.

Free uncombined elements carry an oxidation state of zero, so sulfur in

S_8

is

0

. In tetraoxosulfate(VI) acid (

H_2SO_4

), setting the neutral molecule oxidation sum to zero yields

2(+1) + S + 4(-2) = 0

, giving

S = +6

.

Adım Adım Çözüm

1

Determine the oxidation state of sulfur in free elemental form (

S_8

).

Oxidation state of sulfur in

S_8 = 0

.

By definition, an element in its free or uncombined state has an oxidation state of zero regardless of its atomicity.

2

Calculate the oxidation state of sulfur in

H_2SO_4

.

Oxidation state of sulfur in

H_2SO_4 = +6

.

Assign

+1

for each hydrogen atom and

-2

for each oxygen atom. Solving

2(+1) + S + 4(-2) = 0

gives

+2 + S - 8 = 0

, hence

S = +6

.

Anahtar Kavram

Assigning oxidation states to free elemental forms and central atoms in polyatomic oxoacids

Soru 3563Soru →

For the interval $0^\circ \le x \le 360^\circ$ , solve the trigonometric equation $2\cos^2 x - \cos x - 1 = 0$ . Which set contains all the solutions for $x$ ?

0^\circ, 120^\circ, 240^\circ, 360^\circ

60^\circ, 120^\circ, 240^\circ, 300^\circ

0^\circ, 120^\circ, 360^\circ

60^\circ, 300^\circ, 360^\circ

Cevabı ve açıklamayı göster

Cevap: $0^\circ, 120^\circ, 240^\circ, 360^\circ$

Cevap

The complete set of solutions is

0^\circ, 120^\circ, 240^\circ, 360^\circ

.

Factoring

2\cos^2 x - \cos x - 1 = 0

gives

(2\cos x + 1)(\cos x - 1) = 0

. Setting the first factor to zero yields

\cos x = -\frac{1}{2}

, which has solutions at

120^\circ

and

240^\circ

in the interval

[0^\circ, 360^\circ]

. Setting the second factor to zero gives

\cos x = 1

, which has solutions at

0^\circ

and

360^\circ

. Combining these yields the set

0^\circ, 120^\circ, 240^\circ, 360^\circ

.

Adım Adım Çözüm

1

Factor the quadratic trigonometric equation

(2\cos x + 1)(\cos x - 1) = 0

Treat

\cos x

as a single variable to simplify into standard quadratic factors.

2

Set each factor to zero to find values for

\cos x

\cos x = 1

or

\cos x = -\frac{1}{2}

Zero-product property requires at least one factor to be zero.

3

Solve for

x

in the interval

0^\circ \le x \le 360^\circ

For

\cos x = 1

:

x = 0^\circ, 360^\circ

. For

\cos x = -\frac{1}{2}

:

x = 180^\circ - 60^\circ = 120^\circ

(Quadrant II) and

x = 180^\circ + 60^\circ = 240^\circ

(Quadrant III).

Cosine is negative in Quadrants II and III, with a reference angle of

60^\circ

.

4

Combine all unique solutions in ascending order

x = 0^\circ, 120^\circ, 240^\circ, 360^\circ

Include all solutions within the given domain boundaries.

Anahtar Kavram

Solving Quadratic Trigonometric Equations

Soru 3564Soru →

The aerodynamic drag force $F$ acting on an object moving through a fluid of density $\rho$ with cross-sectional area $A$ at speed $v$ is modeled by the equation $F = \frac{1}{2} C_d \rho^a A^b v^c$ , where $C_d$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the exponent $c$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The numerical value of the exponent

c

is 2.

By applying the principle of dimensional homogeneity, the base dimension of time on the left side is

\text{T}^{-2}

(from force

[F] = \text{M L T}^{-2}

). On the right side, the only quantity containing time is velocity

[v] = \text{L T}^{-1}

, raised to power

c

, giving

\text{T}^{-c}

. Equating the exponents gives

-2 = -c

, so

c = 2

.

Adım Adım Çözüm

1

Identify the base dimensions of each physical quantity in the given equation.

[F] = \text{M L T}^{-2}

,

[\rho] = \text{M L}^{-3}

,

[A] = \text{L}^2

, and

[v] = \text{L T}^{-1}

.

C_d

is dimensionless (

[C_d] = 1

).

Dimensional homogeneity requires both sides of a physical equation to have identical base dimensions.

2

Substitute the base dimensions into the formula

F = \frac{1}{2} C_d \rho^a A^b v^c

and simplify.

\text{M L T}^{-2} = (\text{M L}^{-3})^a (\text{L}^2)^b (\text{L T}^{-1})^c = \text{M}^a \text{L}^{-3a + 2b + c} \text{T}^{-c}.

Combining powers of base dimensions allows direct comparison of corresponding exponents.

3

Equate the exponent of time (T) on both sides of the dimensional equation.

-2 = -c \implies c = 2.

The exponent of T on the left side is -2, which must equal the exponent of T on the right side (-c).

Anahtar Kavram

Principle of Dimensional Homogeneity

Tahmini Süre:1m 15s

Soru 3565Soru →

An alternating current (AC) series circuit contains a resistor of resistance $R = 30\ \Omega$ , an inductor of inductive reactance $X_L = 80\ \Omega$ , and a capacitor of capacitive reactance $X_C = 40\ \Omega$ . The circuit is connected across an AC supply with a peak voltage of $V_0 = 100\sqrt{2}\text{ V}$ . What is the average power dissipated in the circuit?

120\text{ W}

240\text{ W}

13.3\text{ W}

400\text{ W}

Cevabı ve açıklamayı göster

Cevap: $120\text{ W}$

Cevap

The average electrical power dissipated in the circuit is

120\text{ W}

.

The average power dissipated in an AC series circuit depends solely on the resistive component. First, converting peak voltage

V_0 = 100\sqrt{2}\text{ V}

yields an RMS voltage of

V_{\text{rms}} = 100\text{ V}

. Computing the circuit impedance yields

Z = \sqrt{30^2 + (80 - 40)^2} = 50\ \Omega

. This produces an RMS current of

I_{\text{rms}} = \frac{100}{50} = 2\text{ A}

. Finally, substituting into

P_{\text{avg}} = I_{\text{rms}}^2 R

gives

(2)^2 \times 30 = 120\text{ W}

.

Adım Adım Çözüm

1

Calculate the root-mean-square (RMS) voltage of the AC supply.

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\text{ V}

AC power calculations require RMS voltage values rather than peak voltage values.

2

Calculate the net impedance (

Z

) of the series RLC circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (80 - 40)^2} = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\ \Omega

Resistance and reactances combine quadratically due to phase differences between voltage across components.

3

Calculate the RMS current (

I_{\text{rms}}

) flowing through the circuit.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\text{ V}}{50\ \Omega} = 2\text{ A}

Ohm's law for AC circuits states

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}

.

4

Calculate the average power dissipated in the circuit.

P_{\text{avg}} = I_{\text{rms}}^2 R = (2\text{ A})^2 \times 30\ \Omega = 120\text{ W}

Power is dissipated only in resistive elements, as pure inductors and capacitors store and return energy without net loss.

Anahtar Kavram

Impedance and Power in Series AC Circuits

Soru 3566Soru →

In a resonance tube experiment using a tuning fork of constant frequency, the first two consecutive resonant lengths of the air column above the water level are measured to be $23.5\text{ cm}$ and $73.5\text{ cm}$ respectively. What is the end correction of the tube in centimeters?

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The end correction of the tube is

1.5\text{ cm}

.

In a resonance tube closed at one end by water, consecutive resonances occur when the air column length increases by half a wavelength. Subtracting the first resonant length from the second gives

\frac{\lambda}{2} = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}

, which yields

\lambda = 100.0\text{ cm}

and

\frac{\lambda}{4} = 25.0\text{ cm}

. The first resonance condition accounts for end correction through

L_1 + e = \frac{\lambda}{4}

. Substituting

L_1 = 23.5\text{ cm}

gives

e = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}

.

Adım Adım Çözüm

1

Determine the wavelength using consecutive resonant positions

$\frac{\lambda}{2} = L_2 - L_1 = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}$, so $\lambda = 100.0\text{ cm}$

For a column closed at one end, consecutive resonances occur at intervals of half a wavelength.

2

Calculate the quarter-wavelength value

$\frac{\lambda}{4} = \frac{100.0\text{ cm}}{4} = 25.0\text{ cm}$

The fundamental mode position of the displacement antinode corresponds to a distance of one quarter-wavelength from the closed end.

3

Calculate the end correction

$e = \frac{\lambda}{4} - L_1 = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}$

The effective length for the first resonance includes the physical length plus the end correction.

Anahtar Kavram

End Correction in Resonance Air Columns

Soru 3567Soru →

Which of the following conditions is necessary for total internal reflection to occur when light travels between two media?

Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be greater than the critical angle.

Light must travel from an optically less dense medium to an optically denser medium, and the angle of incidence must be greater than the critical angle.

Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be less than the critical angle.

Light must travel from an optically less dense medium to an optically denser medium, and the angle of incidence must equal zero degrees.

Cevabı ve açıklamayı göster

Cevap: Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be greater than the critical angle.

Cevap

Total internal reflection requires light to travel from an optically denser medium to an optically less dense medium with an angle of incidence greater than the critical angle.

For total internal reflection to happen, two strict conditions must be satisfied: (1) light must travel from a medium with a higher refractive index to one with a lower refractive index, and (2) the angle of incidence must be strictly greater than the critical angle for that boundary.

Adım Adım Çözüm

1

Identify the optical density requirement

Light must attempt to pass from a medium of higher refractive index (denser) into a medium of lower refractive index (less dense) so that the refracted ray bends away from the normal.

Bending away from the normal allows the angle of refraction to reach

90^\circ

at the critical angle.

2

Identify the angle of incidence requirement

The angle of incidence in the denser medium must exceed the critical angle (

i > C

).

When the incidence angle exceeds the critical angle, no refraction is possible and all energy is reflected back into the denser medium.

Anahtar Kavram

Conditions for Total Internal Reflection

Soru 3568Soru →

The perimeter of a rectangular playfield is $28\text{ m}$ and its area is $40\text{ m}^2$ . What is the positive difference, in metres, between its length and width?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The positive difference between the length and width of the playfield is 6 metres.

Formulating the system gives

x + y = 14

and

xy = 40

. Substituting

y = 14 - x

yields the quadratic equation

x^2 - 14x + 40 = 0

, which factors into

(x - 10)(x - 4) = 0

. The dimensions are

10\text{ m}

and

4\text{ m}

, giving a positive difference of

10 - 4 = 6\text{ m}

.

Adım Adım Çözüm

1

Set up linear and quadratic equations for perimeter and area

x + y = 14

and

xy = 40

Perimeter formula is

2(x + y) = 28

which simplifies to

x + y = 14

, and area formula is

xy = 40

.

2

Substitute

y = 14 - x

into the quadratic area equation

x(14 - x) = 40 \implies x^2 - 14x + 40 = 0

Substitution reduces the simultaneous system to a single quadratic equation in terms of

x

.

3

Solve the quadratic equation by factoring

(x - 10)(x - 4) = 0 \implies x = 10 \text{ or } x = 4

The roots of the equation give the dimensions of the rectangle.

4

Calculate the positive difference between the two dimensions

10 - 4 = 6

Subtract the smaller dimension from the larger dimension.

Anahtar Kavram

Solving word problems involving simultaneous linear and quadratic equations

Tahmini Süre:1m 30s

Soru 3569Soru →

A quantity $P$ varies partially as $x$ and partially as the square of $y$ . When $x = 2$ and $y = 3$ , $P = 24$ , and when $x = 5$ and $y = 1$ , $P = 17$ . What is the value of $P$ when $x = 4$ and $y = 3$ ?

30

35

26

21

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The value of

P

is 30.

The relationship follows the partial variation formula

P = k_1 x + k_2 y^2

. Substituting the given conditions gives

2k_1 + 9k_2 = 24

and

5k_1 + k_2 = 17

. Solving these simultaneous equations yields

k_1 = 3

and

k_2 = 2

. Evaluating

P = 3(4) + 2(3^2)

produces

12 + 18 = 30

.

Adım Adım Çözüm

1

Set up the general formula for partial variation.

P = k_1 x + k_2 y^2

, where

k_1

and

k_2

are constants.

Partial variation combines terms linearly with separate variation constants.

2

Substitute the given pairs of values to form simultaneous linear equations.

Equation (1):

2k_1 + 9k_2 = 24

; Equation (2):

5k_1 + k_2 = 17

.

Plugging in

(x=2, y=3, P=24)

and

(x=5, y=1, P=17)

creates a system of equations in terms of

k_1

and

k_2

.

3

Solve the simultaneous linear equations for

k_1

and

k_2

.

From Equation (2),

k_2 = 17 - 5k_1

. Substitute into Equation (1):

2k_1 + 9(17 - 5k_1) = 24 \implies -43k_1 = -129 \implies k_1 = 3

. Then

k_2 = 17 - 5(3) = 2

.

Finding the specific values of the variation constants is required to complete the formula.

4

Calculate

P

for

x = 4

and

y = 3

using the complete formula

P = 3x + 2y^2

.

P = 3(4) + 2(3^2) = 12 + 2(9) = 12 + 18 = 30

.

Evaluating the relationship with the target parameters produces the final answer.

Anahtar Kavram

Partial Variation and Simultaneous Linear Equations

Tahmini Süre:2m 0s

Soru 3570Soru →

Given the $3 \times 3$ matrix $A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{pmatrix}$ , find the positive value of $x$ for which $\det(A) = -2$ .

Cevabı ve açıklamayı göster

Cevap: 2.5

Cevap

The positive value of

x

is 2.5.

Expanding the determinant of matrix

A

along the third row gives

1(4 - 1) - x(2x - 3) = -2x^2 + 3x + 3

. Setting this equal to

-2

yields

2x^2 - 3x - 5 = 0

. Factoring gives

(2x - 5)(x + 1) = 0

, yielding solutions

x = 2.5

and

x = -1

. Taking the positive value gives

x = 2.5

.

Adım Adım Çözüm

1

Calculate the determinant of matrix

A

in terms of

x

\det(A) = -2x^2 + 3x + 3

Expanding along the third row simplifies computation because of the zero entry.

2

Set the determinant expression equal to

-2

and rearrange terms

2x^2 - 3x - 5 = 0

Setting

-2x^2 + 3x + 3 = -2

forms a standard quadratic equation.

3

Factorize the quadratic equation to find the roots

x = 2.5

or

x = -1

Factoring

(2x - 5)(x + 1) = 0

yields two real solutions.

4

Filter for the positive value requested in the stem

x = 2.5

The question specifically requires the positive value of

x

.

Anahtar Kavram

Determinant of a 3x3 Matrix and Quadratic Equation Solving

Tahmini Süre:2m 30s

Soru 3571Soru →

If $\log_3 x - 2 \log_x 3 = 1$ , what is the sum of all valid solutions for $x$ ?

\frac{28}{3}

12

1

9

Cevabı ve açıklamayı göster

Cevap: $\frac{28}{3}$

Cevap

The sum of all valid solutions for

x

is

\frac{28}{3}

.

By using the change of base identity

\log_x 3 = \frac{1}{\log_3 x}

and substituting

u = \log_3 x

, the given equation simplifies to

u - \frac{2}{u} = 1

. Multiplying by

u

gives the quadratic equation

u^2 - u - 2 = 0

, which factors as

(u - 2)(u + 1) = 0

. This yields

u = 2

or

u = -1

. Converting back to

x

gives

x = 3^2 = 9

and

x = 3^{-1} = \frac{1}{3}

. Both values satisfy domain constraints for logarithmic base (

x > 0

and

x \neq 1

). Adding these values gives

9 + \frac{1}{3} = \frac{28}{3}

.

Adım Adım Çözüm

1

Apply the change of base formula to express the equation with a common logarithmic base

Using

\log_x 3 = \frac{1}{\log_3 x}

, the equation becomes

\log_3 x - \frac{2}{\log_3 x} = 1

.

Change of base allows all terms containing

x

to be represented in terms of a single logarithmic expression.

2

Substitute

u = \log_3 x

and clear the fraction to form a quadratic equation

u - \frac{2}{u} = 1 \implies u^2 - u - 2 = 0

.

Multiplying through by

u

(where

u \neq 0

) transforms the logarithmic relationship into a standard quadratic algebraic equation.

3

Factor the quadratic equation to find the values of

u

(u - 2)(u + 1) = 0 \implies u = 2 \text{ or } u = -1

.

Factoring isolates the linear roots for the substituted variable

u

.

4

Solve for

x

from

u = \log_3 x

and verify validity within the domain

For

u = 2

:

\log_3 x = 2 \implies x = 3^2 = 9

.
For

u = -1

:

\log_3 x = -1 \implies x = 3^{-1} = \frac{1}{3}

.
Both solutions are positive and

x \neq 1

, so both are valid.

Converting from logarithmic form to exponential form retrieves the original variable

x

.

5

Calculate the sum of all valid solutions

Sum =

9 + \frac{1}{3} = \frac{27}{3} + \frac{1}{3} = \frac{28}{3}

.

The question requires finding the total sum of all permissible real solutions for

x

.

Anahtar Kavram

Logarithmic Change of Base and Equations Reducible to Quadratics

Soru 3572Soru →

The table below shows the distribution of quiz scores for a class of students:

Score Interval	Frequency ( $f$ )
$1 - 5$	$3$
$6 - 10$	$5$
$11 - 15$	$7$
$16 - 20$	$5$

Calculate the mean score of the distribution.

Cevabı ve açıklamayı göster

Cevap: 11.5

Cevap

The mean score of the distribution is 11.5.

The mean for a grouped frequency table is calculated by taking the sum of the products of each midpoint (

x

) and its frequency (

f

), divided by the sum of all frequencies (

f

). Here,

\sum fx = 230

and

\sum f = 20

, yielding

\bar{x} = 11.5

.

Adım Adım Çözüm

1

Determine the class midpoints (

x

) for each interval.

Midpoints are 3, 8, 13, and 18.

Midpoints serve as the representative values for each class interval.

2

Calculate the product of each midpoint and frequency (

fx

).

Products are 9, 40, 91, and 90.

To find the total contribution of each class interval.

3

Find the sum of all frequencies (

\sum f

) and products (

\sum fx

).

\sum f = 20

and

\sum fx = 230

.

These totals are required for the grouped mean formula.

4

Divide the total product sum by total frequency.

\bar{x} = \frac{230}{20} = 11.5

.

Applying the formula for the mean of grouped data

\bar{x} = \frac{\sum fx}{\sum f}

.

Anahtar Kavram

Mean of Grouped Data using Class Midpoints

Soru 3573Soru →

Find the smallest non-negative integer $x$ that satisfies the linear modular congruence $3x \equiv 8 \pmod{11}$ .

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The smallest non-negative integer

x

is 10.

Evaluating

3x \equiv 8 \pmod{11}

by testing multiples of

11

added to

8

gives

30

, which divided by

3

yields

x = 10

. Since

10 \in [0, 10]

, it is the canonical solution.

Adım Adım Çözüm

1

Convert the modular congruence into an algebraic equation

3x = 8 + 11k

for an integer

k

By definition of congruence modulo

11

,

3x - 8

must be a multiple of

11

.

2

Find the smallest integer

k \ge 0

such that

8 + 11k

is divisible by 3

When

k = 2

,

8 + 11(2) = 30

30

is divisible by

3

(

30 / 3 = 10

).

3

Divide by 3 to isolate

x

x = 10

3(10) = 30 \equiv 8 \pmod{11}

.

Anahtar Kavram

Linear Modular Congruence

Tahmini Süre:1m 15s

Soru 3574Soru →

Find the real value of $x$ that satisfies the exponential equation $4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$ .

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The value of

x

is

1.5

(or

\frac{3}{2}

).

By using index laws to group base-2 terms on the left side and base-3 terms on the right side, we obtain

2^{2x}\left(\frac{3}{2}\right) = 3^x\left(\frac{4}{\sqrt{3}}\right)

. Rearranging gives

\left(\frac{4}{3}\right)^x = \frac{8}{3\sqrt{3}} = \left(\frac{4}{3}\right)^{\frac{3}{2}}

. Equating indices gives

x = 1.5

.

Adım Adım Çözüm

1

Group like exponential terms with base 2 and base 3 on opposite sides of the equation.

4^x + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}

Grouping terms with common prime bases allows for factoring exponential terms.

2

Apply the product and power laws of indices:

4^x = 2^{2x}

,

2^{2x-1} = 2^{-1} \cdot 2^{2x}

,

3^{x \pm \frac{1}{2}} = 3^x \cdot 3^{\pm \frac{1}{2}}

.

2^{2x} + \frac{1}{2} \cdot 2^{2x} = 3^x \cdot \sqrt{3} + 3^x \cdot \frac{1}{\sqrt{3}}

Separating the variable exponents from constant exponents prepares each side for factoring.

3

Factor out

2^{2x}

from the left side and

3^x

from the right side, then simplify arithmetic terms.

2^{2x}\left(\frac{3}{2}\right) = 3^x\left(\frac{4}{\sqrt{3}}\right)

Factoring isolates the variable terms

2^{2x}

and

3^x

from numerical constants.

4

Divide to form the ratio

\frac{4^x}{3^x} = \left(\frac{4}{3}\right)^x

and simplify the numerical fraction on the right.

\left(\frac{4}{3}\right)^x = \frac{8}{3\sqrt{3}}

Expressing both sides with unified variable bases facilitates solving for

x

by equating powers.

5

Rewrite

\frac{8}{3\sqrt{3}}

as a power of

\frac{4}{3}

and solve for

x

.

\left(\frac{4}{3}\right)^x = \left(\frac{4}{3}\right)^{\frac{3}{2}} \implies x = \frac{3}{2} = 1.5

Since

\frac{8}{3\sqrt{3}} = \frac{4^{3/2}}{3^{3/2}} = (\frac{4}{3})^{3/2}

, equating exponents yields

x = 1.5

.

Anahtar Kavram

Solving mixed-base exponential equations by grouping, factoring, and converting to a unified base ratio.

Tahmini Süre:3m 0s

Soru 3575Soru →

Match each vocabulary word in the left column with its precise antonym (word opposite in meaning) in the right column.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Diligent pairs with Lazy; Humble pairs with Arrogant; Friendly pairs with Hostile; Permanent pairs with Temporary.

Each word on the left is matched with its exact antonym on the right: Diligent (hardworking) is opposite to Lazy; Humble (modest) is opposite to Arrogant; Friendly (kind) is opposite to Hostile; Permanent (lasting) is opposite to Temporary.

Adım Adım Çözüm

1

Define each target word in the left column.

Diligent means hardworking; Humble means modest; Friendly means pleasant/kind; Permanent means lasting indefinitely.

Establishing clear definitions is necessary to determine precise antonyms.

2

Find the corresponding word in the right column that expresses the direct opposite meaning.

Diligent matches Lazy; Humble matches Arrogant; Friendly matches Hostile; Permanent matches Temporary.

Antonyms express contrasting or opposing semantic qualities.

Anahtar Kavram

Antonyms and Words Opposite in Meaning

Soru 3576Soru →

Solve the simultaneous equations $y = 2x + 1$ and $y = x^2 - 2$ . Which of the following represents the complete set of solution pairs $(x, y)$ ?

(3, 7)

and

(-1, -1)

(-3, -5)

and

(1, 3)

(3, 5)

and

(-1, -3)

(3, 7)

and

(1, 3)

Cevabı ve açıklamayı göster

Cevap: $(3, 7)$ and $(-1, -1)$

Cevap

The complete set of solution pairs

(x, y)

is

(3, 7)

and

(-1, -1)

.

Equating

2x + 1 = x^2 - 2

yields

x^2 - 2x - 3 = 0

. Factoring gives

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

, leading to

x = 3

or

x = -1

. Substituting these

x

-values into

y = 2x + 1

gives

y = 7

for

x = 3

, and

y = -1

for

x = -1

. Thus, the solution pairs are

(3, 7)

and

(-1, -1)

.

Adım Adım Çözüm

1

Equate the linear expression for

y

to the quadratic expression for

y

.

2x + 1 = x^2 - 2

Since both expressions equal

y

, setting them equal eliminates

y

.

2

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 2x - 3 = 0

Subtract

2x

and

1

from both sides.

3

Factor the quadratic equation to find the values of

x

.

(x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1

Determine two numbers that multiply to

-3

and add to

-2

.

4

Substitute each

x

-value back into the linear equation

y = 2x + 1

to find the corresponding

y

-value.

For

x = 3

,

y = 2(3) + 1 = 7

. For

x = -1

,

y = 2(-1) + 1 = -1

.

Calculate the exact coordinate pairs

(x, y)

that satisfy both equations.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using algebraic substitution.

Soru 3577Soru →

A particle starts from rest and accelerates uniformly at a rate of $4\text{ m/s}^2$ for a duration $t_1$ . Immediately after reaching its maximum velocity, it decelerates uniformly at $2\text{ m/s}^2$ until coming to rest. If the total distance covered during the entire motion is $600\text{ m}$ , what is the total time of motion in seconds?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The total time of motion is 30 seconds.

For a two-stage motion starting and ending at rest, the peak velocity is

v_{\text{max}} = a_1 t_1 = a_2 t_2

, giving a time ratio

t_2 / t_1 = a_1 / a_2 = 4 / 2 = 2

. The total distance

S

is the area under the velocity-time triangle,

S = \frac{1}{2} v_{\text{max}} (t_1 + t_2) = \frac{1}{2} (4 t_1) (3 t_1) = 6 t_1^2

. Setting

6 t_1^2 = 600

yields

t_1 = 10\text{ s}

, which gives a total time

T = t_1 + t_2 = 30\text{ s}

.

Adım Adım Çözüm

1

Relate maximum velocity to the acceleration time

t_1

v_{\text{max}} = 4 t_1

Using

v = u + a t

starting from rest (

u = 0

).

2

Relate deceleration time

t_2

to

t_1

t_2 = 2 t_1

The final velocity is

0

, so

0 = v_{\text{max}} - a_2 t_2 \implies t_2 = \frac{4 t_1}{2} = 2 t_1

.

3

Express the total displacement

S

as a function of

t_1

S = 6 t_1^2

Displacement during acceleration

s_1 = \frac{1}{2}(4)t_1^2 = 2 t_1^2

. Displacement during deceleration

s_2 = \frac{1}{2}(2)(2 t_1)^2 = 4 t_1^2

. Total

S = 2 t_1^2 + 4 t_1^2 = 6 t_1^2

.

4

Solve for the acceleration time

t_1

t_1 = 10\text{ s}

Given

S = 600\text{ m}

, we have

6 t_1^2 = 600 \implies t_1^2 = 100 \implies t_1 = 10\text{ s}

.

5

Calculate the total time of motion

T

T = 30\text{ s}

Total time is the sum of both phases:

T = t_1 + t_2 = t_1 + 2 t_1 = 3 t_1 = 3(10) = 30\text{ s}

.

Anahtar Kavram

Multi-stage uniform motion and average velocity relations

Tahmini Süre:3m 0s

Soru 3578Soru →

A curve is defined by the equation $y = x^2 - 5x + 6$ . What is the $x$ -intercept of the line normal to the curve at the point where $x = 1$ ?

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

The x-intercept of the normal line to the curve at x = 1 is -5.

Substituting

x = 1

into

y = x^2 - 5x + 6

yields

y = 2

, identifying the point

(1, 2)

. Differentiating gives

\frac{dy}{dx} = 2x - 5

, which equals

-3

at

x = 1

. The normal line gradient is the negative reciprocal,

\frac{1}{3}

. The line equation

y - 2 = \frac{1}{3}(x - 1)

simplifies to

x - 3y + 5 = 0

. Setting

y = 0

gives

x = -5

.

Adım Adım Çözüm

1

Calculate the y-coordinate at x = 1 to determine the point of tangency

Substituting

x = 1

into

y = x^2 - 5x + 6

gives

y = (1)^2 - 5(1) + 6 = 2

, yielding the point

(1, 2)

.

The normal line intersects the curve at the point of tangency.

2

Find the derivative of the curve and evaluate the tangent slope

\frac{dy}{dx} = 2x - 5

. At

x = 1

,

m_t = 2(1) - 5 = -3

.

The derivative evaluated at a specific point gives the slope of the tangent line to the curve.

3

Compute the slope of the normal line

m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}

.

The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent gradient.

4

Construct the normal line equation and determine its x-intercept

Using point-slope form:

y - 2 = \frac{1}{3}(x - 1) \implies 3y - 6 = x - 1 \implies x - 3y + 5 = 0

. Setting

y = 0

gives

x + 5 = 0 \implies x = -5

.

The

x

-intercept occurs where the line crosses the

x

-axis, meaning

y = 0

.

Anahtar Kavram

The slope of the normal line to a curve

y = f(x)

at

x = a

is the negative reciprocal of the derivative at that point,

m_n = -\frac{1}{f'(a)}

.

Soru 3579Soru →

If the matrix $P = \begin{pmatrix} k & 3 & 1 \\ 2 & 1 & k \\ 4 & 2 & 0 \end{pmatrix}$ is singular, find the non-zero value of $k$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The non-zero value of

k

is 6.

For matrix P to be singular, its determinant must be 0. Expanding along row 3 yields 4(3k - 1) - 2(k^2 - 2) = 12k - 4 - 2k^2 + 4 = -2k^2 + 12k = 0. Factoring gives -2k(k - 6) = 0, which yields k = 0 or k = 6. The non-zero value is 6.

Adım Adım Çözüm

1

Set the determinant of matrix P to 0

\det(P) = 0

A matrix is singular if and only if its determinant equals zero.

2

Evaluate the 3x3 determinant by expanding along the third row

4 \cdot (3k - 1) - 2 \cdot (k^2 - 2) + 0 = 0

Expanding along the third row takes advantage of the zero entry to simplify computation.

3

Expand and combine like terms

-2k^2 + 12k = 0

12k - 4 - 2k^2 + 4 reduces to -2k^2 + 12k.

4

Factor out common factors and solve for k

-2k(k - 6) = 0 \implies k = 0 \text{ or } k = 6

Applying the zero-product property.

5

Select the required root

k = 6

The question specifies the non-zero value of k.

Anahtar Kavram

Determinant of a 3x3 matrix and singular matrix condition

Tahmini Süre:2m 0s

Soru 3580Soru →

A body of mass $0.4\text{ kg}$ suspended vertically from a helical spring produces a static extension of $0.1\text{ m}$ . The body is then pulled down further and set into vertical simple harmonic motion with an amplitude of $0.05\text{ m}$ . What is the maximum velocity of the body in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

The maximum velocity of the body during oscillation is

0.5\text{ m/s}

.

At static equilibrium, weight balances restoring force (

mg = ke

), giving

\frac{k}{m} = \frac{g}{e} = \frac{10}{0.1} = 100\text{ s}^{-2}

. The angular frequency is

\omega = \sqrt{\frac{k}{m}} = 10\text{ rad/s}

. In SHM, the maximum velocity occurs at the central equilibrium position and is given by

v_{\max} = \omega A = 10 \times 0.05 = 0.5\text{ m/s}

.

Adım Adım Çözüm

1

Relate spring stiffness to static extension

\frac{k}{m} = 100\text{ s}^{-2}

At vertical static equilibrium, the weight of the mass equals the restoring force:

mg = ke \implies \frac{k}{m} = \frac{g}{e} = \frac{10\text{ m/s}^2}{0.1\text{ m}} = 100\text{ s}^{-2}

.

2

Determine the angular frequency

\omega = 10\text{ rad/s}

The angular frequency of a mass-spring system is given by

\omega = \sqrt{\frac{k}{m}} = \sqrt{100} = 10\text{ rad/s}

.

3

Calculate maximum velocity

v_{\max} = 0.5\text{ m/s}

The maximum speed in simple harmonic motion occurs at the equilibrium position and is computed using

v_{\max} = \omega A = 10\text{ rad/s} \times 0.05\text{ m} = 0.5\text{ m/s}

.

Anahtar Kavram

Maximum velocity and angular frequency derived from static extension in Simple Harmonic Motion

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