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3 years ago

The area of the base of a prism is 21 cm2. The perimeter of the base is 20 cm. The height of the prism is 8 cm.

1 answer:

7 0

The lateral area of the prism equals to height * perimeter of the base. So the lateral area is 160 cm2. And the surface area of the prism equals to the area of the base*2+the lateral area= 202 cm2.

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Kim rode her bicycle 135 miles in 9 weeks, riding the same distance each week. Eric rode his bicycle 102 miles in 6 weeks, ridin

Juli2301 [7.4K]

Answer:

Eric rode 2 more miles per week than Kim rode.

Step-by-step explanation:

Find how many miles per week Kim rode by dividing the number of miles she rode by the number of weeks:

135/9

= 15

Find how many miles per week Eric rode:

102/6

= 17

So, Kim rode 15 miles per week and Eric rode 17 miles per week.

The answer is Eric rode 2 more miles per week than Kim rode.

4 0

3 years ago

Simplify 144 square root.

Brrunno [24]

Answer:

sqrt(144) = ±12

Step-by-step explanation:

sqrt(144)

What number, when multiplied by itself, gives 144

12*12 = 144

-12*-12 = 144

sqrt(144) = ±12

4 0

2 years ago

Read 2 more answers

The length of a rectangle is 3 meters more than twice its width.The perimeter of the rectangle is 48 meters. Let W represent the

lord [1]

Perimeter of a rectangle: 2(l+w)

Let width be 'x'

Length is 3 meters more than twice 'x'
Length= 2x+3.

L= x
W = 2x+3

Formula =
2(x+2x+3)
2(3x+3)
6x+6.
6(x+1)

8 0

3 years ago

(07.05) Graph the first six terms of a sequence where a = -10 and d = 3.

butalik [34]

Answer:

Step-by-step explanation:

Using the formular, Sn = n/2(2a + (n - 1)d)

where, n = 6: Sn = 6/2( -20 + 5(3)) = 3(15 - 20)

∴ Sn = 15

7 0

3 years ago

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$