All categories

3 years ago

Match each function with the corresponding function formula when h(x) = 5 - 3x and g(x) = -3 x + 5.

Mathematics

1 answer:

laila [671]3 years ago

6 0

Answer:

To match each functions with the corresponding function formula when h(x) = 5 - 3x and g(x) = -3 x + 5.

1. k(x) = (3h - 5g)(x)

= = 3(5 - 3x)-5(-3 x + 5)

= = 15 - 9x + 15x - 25

K(x) = -10 + 6x

2. k(x) = (h - g)(x)

= = (5 - 3x) - (-3 x + 5)

= = 0

k(x) = 0

3. k(x) = (5g + 3h)(x)

= = 5(-3 x + 5)+3(5 - 3x)

= = -15x + 25 + 15 - 9x

k(x) = - 24x +40

4. k(x) = (3g + 5h)(x)

= = 3 (-3 x + 5) + 5(5 - 3x)

= = -9x + 15 +25 -15x

k(x) = -24x + 40

5. k(x) = (g + h)(x)

= = (-3 x + 5) + (5 - 3x)

k(x) = -6x + 10

6. k(x) = (5h - 3g)(x)

= = 5(5 - 3x) - 3(-3 x + 5)

= =25 - 15x + 9x - 15

k(x) = 10 - 6x

You might be interested in

135% as a fraction or a mixed number fully simplified please 50 POINTS + BRAINILEST

Valentin [98]

Answer:

27/20

Step-by-step explanation:

135% = 135/100 = (27/20 × 5) = 27/20 = 1 7/20 I think

4 0

2 years ago

Identify which of these designs is most appropriate for the given experiment: completely randomized design, randomized block

Eduardwww [97]

Answer:

Completely Randomized Design

Step-by-step explanation:

In a completely randomized design, the samples are randomly assigned to the treatment without creating any blocks or groups.

Like here, in the given scenario, we do not have to divide subjects in two groups as they are all same.

Whereas, in a randomized block design, the participants are divided into subgroups in a way, that the variability within the blocks is less than the variability between blocks.

After dividing, the participants within each block are randomly assigned to treatment conditions.

Hence, the completely randomized design is used here.

3 0

2 years ago

Which two equations can you use to find the adult and child admission fees for a puppet show? Let a represent the fee for adults

vlabodo [156]

$45a + 30c = 315 \\ 50a + 40c = 370$
The answers are C and E

Hope this helps. - M

3 0

3 years ago

Read 2 more answers

A tank contains 180 gallons of water and 15 oz of salt. water containing a salt concentration of 17(1+15sint) oz/gal flows into

Stels [109]

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

$\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}$

and flows out at a rate of

$\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}$

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

$\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))$

Multiply both sides by the integrating factor, $e^{t/180}$ , and rewrite the left side as the derivative of a product.

$e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))$

$\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))$

Integrate both sides with respect to t (integrate the right side by parts):

$\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt$

$\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C$

Solve for A(t) :

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}$

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

$\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}$

So,

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}$

Recall the angle-sum identity for cosine:

$R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)$

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

$R \cos(\theta) = -\dfrac{66,096,000}{32,401}$

$R \sin(\theta) = \dfrac{367,200}{32,401}$

Recall the Pythagorean identity and definition of tangent,

$\cos^2(x) + \sin^2(x) = 1$

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$

Then

$R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}$

and

$\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)$

so we can rewrite A(t) as

$\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}$

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

$24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}$

and

$24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}$

which is to say, with amplitude

$2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}$

6 0

2 years ago

How do you divide fractions

Sergeeva-Olga [200]

Leave the first fraction in the equation alone.
Turn the division sign into a multiplication sign.
Flip the second fraction over (find its reciprocal).
Multiply the numerators (top numbers) of the two fractions together. ...
Multiply the denominators (bottom numbers) of the two fractions together.

6 0

2 years ago