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3 years ago

Find the surface area of the prism. 5cm 7cm 4cm

Mathematics

1 answer:

Georgia [21]3 years ago

4 0

To find the volume it is Length x Width x Height
so 7 x 5 x 4 = 140cm
140 cm is the answer
Hoped this helped =)

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The students in Mr. Nolan's class are writing expressions for the perimeter of a rectangle of side length ? and width w.

uranmaximum [27]

You wanna know witch student is correct?

6 0

3 years ago

Given f(x) = √x and g(x) = x+2 find the domain of f(g(x)).

REY [17]

Answer:

E) None of the above

Step-by-step explanation:

Given $f(x)=\sqrt{x}$ and $g(x)=x+2$ , then $f(g(x))=\sqrt{x+2}$ . This shifts the graph of the parent function $f(x)=\sqrt{x}$ 2 units to the left. Thus, the domain of the function is $[-2,\infty)$ , making none of the above true.

8 0

2 years ago

HELP!! PLEASE!!! THANK YOU!!

Blizzard [7]

Answer:

C. 8x - 3y = 5

Step-by-step explanation:

7 - 3(x - y) = 5x + 2

Standard form: Ax + By = C

7 -3x + 3y = 5x + 2

7 -8x + 3y = 2

-8x + 3y = -5

Divide by -1 to each number

8x - 3y = 5

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

3 years ago

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

iren [92.7K]

9514 1404 393

Answer:

32 feet

Step-by-step explanation:

0 seconds after launch, the rocket is still sitting on the tower. Its height is...

y = -16(0^2) +57(0) +32

y = 0 + 0 + 32 = 32

The rocket was launched from a tower 32 feet high.

6 0

3 years ago