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4 years ago

Please help with this math question

Mathematics

1 answer:

cestrela7 [59]4 years ago

4 0

The answer would be D because if you notice the ratios between coordinates and their placement, the first transformation divides the coordinates by 8, or dilating it by a scale factor of 1/8. Then, in the second transformation, each of the coordinates' x-values increased by four, so it moves over four spaces to the right.

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Two children want to race around the city block. The length of the block is 80 yards. The width of the block is 60 yards.

kumpel [21]

I thank it is a. yards i am no sure that is right

7 0

3 years ago

Read 2 more answers

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

Whoevr answers this question correctly gets brainly<br><br>and show your work please

blsea [12.9K]

Answer:

1/4

Step-by-step explanation:

The chance of it landing heads up once would be 1/2

The chance of it landing heads up twice would be 1/4

The chance of it landing heads up three times would be 1/8

The chance of it landing heads up four times would be 1/16

See the pattern?

Pattern: <em><u>Divide the denominator by 2.</u></em>

<h2><em>Please give brainliest!!! :) </em></h2>

3 0

3 years ago

-6x^3-y^2-3 ×=2 y=4 what is value

harkovskaia [24]

- 67

substitute the given values for x and y into the expression and evaluate

= - 6(2)³ - (4)² - 3 = ( - 6 × 8 ) - 16 - 3 = - 48 - 16 - 3 = - 67

5 0

3 years ago

Read 2 more answers

Evaluate the integral using integration by parts with the indicated choices of u and dv. (Use C for the constant of integration.

trasher [3.6K]

Answer:

$\frac{xe^{7x}}{7} + \frac{e^{7x}}{49}$

Step-by-step explanation:

Given the integral equation

$\int\limits{xe^{7x}} \, dx \\$

According to integration by part;

$\int\limits {u} \, dv = uv + \int\limits {v} \, du$

u = x, dv = e^7x

du/dx = 1

du = dx

$v = \int\limits {e^{7x}} \, dx \\v = e^7x/7$

Substitute the given values into the formula;

$\int\limits {xe^{7x}} \, dx = x(e^{7x}/7) + \int\limits ({e^{7x}/7}) \, dx\\\int\limits {xe^{7x}} \, dx = \frac{xe^{7x}}{7} + \frac{e^{7x}}{7*7} \\\int\limits {xe^{7x}} \, dx = \frac{xe^{7x}}{7} + \frac{e^{7x}}{49}$

3 0

3 years ago