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3 years ago

Lorde help me surface area 10 points

Mathematics

1 answer:

USPshnik [31]3 years ago

6 0

Answer:

1560 in^2

Step-by-step explanation:

10x22 = 220 bottom rectangle

1/2(10)(24) = 120 side triangle

22x26 = 572 front rectangle

24x22 = 528 back rectangle

220+120+120-572+528= 1560

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Find the nth term for the sequence 8, 14, 20, 26

harina [27]

It adds six every time so the answer is 56

4 0

3 years ago

enot [183]

Answer:

y = 5/2x

y = 35/2

Step-by-step explanation:

direct variation

y = kx

Substitute the values for x and y

15 = k(6)

Divide each side by 6 yo find k

15/6 = k

5/2 = k

The equation is y = 5/2 x

Now we know x =7

y = 5/2(7)

y = 35/2

8 0

4 years ago

Read 2 more answers

Jenna gave 5 grapes to each of her sisters she has 3 sisters what is the total number of grapes Jenna gave to her sisters ,a 3 b

Lera25 [3.4K]

The answer is D she's giving 5 grapes to each of her sisters so 5 times 3 = 15

7 0

3 years ago

On a test with a population mean of 75 and standard deviation equal to 16, if the scores are normally distributed, what percenta

sashaice [31]

Answer:

Percentage of scores that fall between 70 and 80 = 24.34%

Step-by-step explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N( $\mu,\sigma^{2}$ )

The z probability is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1) where, $\mu$ = 75 and $\sigma$ = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P( $\frac{X-\mu}{\sigma}$ < $\frac{80-75}{16}$ ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P( $\frac{X-\mu}{\sigma}$ < $\frac{70-75}{16}$ ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

= 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%

6 0

4 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago