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3 years ago

URGENT!! Really need help on this test

Mathematics

2 answers:

daser333 [38]3 years ago

7 0

Answer:

Step-by-step explanation:

Since the triangles are congruent

guajiro [1.7K]3 years ago

3 0

Answer:

angle S is 38 this is right answer

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Which expression is equivalent to 7a^2b + 10a^2b^2 + 14a^2b^3?

vladimir1956 [14]

7a^2b + 10a^2b^2 + 14a^2b^3 =
a^2b (7 + 10b + 14b^2) <==

3 0

3 years ago

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5x^{2} + 3x = x^{2} + 7x

Ipatiy [6.2K]

$5x^{2} + 3x = x^{2} + 7x\\
4x^2-4x=0\\
4x(x-1)=0\\
x=0 \vee x=1$

8 0

3 years ago

Points Possible: 1, Points Correct: 0 Given the equation y = x + 9, what is the y -intercept?

-Dominant- [34]

Answer:

Step-by-step explanation:

The equation says y = x + 9

Therefore, y = 9 since x = 0 on the y axis.

Picture of graph also attached :)

8 0

3 years ago

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.

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2 years ago

Two urns contain white balls and yellow balls. The first urn contains 2 white balls and 7 yellow balls and the second urn contai

marissa [1.9K]

P(WW)=(2/9)(10/14)

P(WW)=10/63

3 0

3 years ago

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