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2 years ago

Find the area of the figure. (Sides meet at right angles.)

Mathematics

1 answer:

son4ous [18]2 years ago

7 0

Answer:

40 yd^2

Step-by-step explanation:

Add the square at the bottom to make one large rectangle. Subtract the square not included from that area to get the figure shown (see attachment).

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Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Which expression is equivalent to (2g^3+4)^2

babunello [35]

(2g^3+4)^2= 4g^6 + 16g^3 + 16
The answer is B.

7 0

3 years ago

Read 2 more answers

I’m not sure need help! ANSWER QUICKKK PLZZ

Vadim26 [7]

Option D, 5^9 is the correct answer

4 0

3 years ago

There are 1,657 souvenir paperweights that need to be packed in boxes. Each box will hold 17

blondinia [14]

Answer:

Therefore we can say that 98 boxes will be needed to hold all the paperweights.

Step-by-step explanation:

i) there are 1657 souvenir paperweights that need to be packed in boxes.

ii) each box will hold 17 paperweights

iii) therefore the number of boxes that will be needed are

= $\dfrac{total\hspace{0.15cm} number\hspace{0.15cm} souvenir\hspace{0.15cm} paperweights}{number\hspace{0.15cm} of\hspace{0.15cm} paperweights\hspace{0.15cm} per\hspace{0.15cm} box}$ = $\dfrac{1657}{17}$ = 97.471

iv) Therefore we can say that 98 boxes will be needed to hold all the paperweights.

4 0

3 years ago

HELP ASAP!! WILL MARK AS BRAINLIEST IF ANSWERED NOW!!!!

tiny-mole [99]

Answer: She should blend 98 lbs of high-quality beans.

She should blend 72 lbs of cheaper beans

Step-by-step explanation:

Let x represent the number of pounds of high quality beans that she should blend.

Let y represent the number of pounds of cheaper beans that she should blend.

She needs to prepare 170 lbs of blended coffee beans. This means that

x + y = 170

She plans to do this by blending together a high-quality bean costing $4.75 per pound and a cheaper bean at $2.00 per pound. The blend would sell for $3.59 per pound. This means that the total cost of the blend would be 3.59×170 = $610.3. This means that

4.75x + 2y = 610.3 - - - - - - - - - -1

Substituting x = 170 - y into equation 1, it becomes

4.75(170 - y) + 2y = 610.3

807.5 - 4.75y + 2y = 610.3

- 4.75y + 2y = 610.3 - 807.5

- 2.75y = - 197.2

y = - 197.2/-2.75 = 71.9

y = 72 pounds

x = 170 - y = 170 - 71.9

x = 98.1

x = 98 pounds

4 0

3 years ago