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3 years ago

I need answers pls I have a test right now

Mathematics

1 answer:

barxatty [35]3 years ago

6 0

On question 9, x = 0.46284330

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6. A prism has bases that are equilateral triangles with sides lengths of 10 inches and a length of 30 inches.

eimsori [14]

Answer:

The volume of the prism is $1,299\ in^{3}$

Step-by-step explanation:

we know that

The volume of the prism is equal to

$V=BL$

where

B is the area of the triangular base

L is the length of the prism

we have

$L=30\ in$

Find the area of the base B

The area of a equilateral triangle is equal to

$B=\frac{1}{2}(10)^{2} sin(60\°)$

$B=25\sqrt{3}\ in^{2}$

substitute

$V=(25\sqrt{3})(30)=1,299\ in^{3}$

8 0

3 years ago

A. g(x)=3x^2 b. g(x)=(1/9x)^2 c. g(x)-1/3x^2 d. g(x)(1/3x)2

zaharov [31]

Answer:

the awnser would be a(g(x)=3x2

6 0

3 years ago

6. I think the answer is H, could you help and check?!

Sliva [168]

Set up an equation to solve.
70% of original price equals $550, the amount Susan is paying.
0.70x = 550

8 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

Which of the following tables is a relation that represents a function?

uysha [10]

The first one is not a function due to the rule that there can not be more than one x, such as there is a repeated number 1 , 2 , 1 , 4 There should not be two
x = 1 terms.

Same the same rule applies to the second option as well as the last.

Your answer is C. or the third option
x= 6, 5, 4, 1
y=6, 4, 6, 2

Hope this Helps

3 0

3 years ago