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3 years ago

Can someone help me with Surface area?

Mathematics

1 answer:

postnew [5]3 years ago

8 0

Find the area of each individual shape and add those answers to get the entire sum

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4 glue sticks cost $7.76. Write a proportion

damaskus [11]

1.94

Divide 7.76 by 4 to get the cost of each glue stick

3 0

2 years ago

A painter works 37.5 hours one week. If she worked 5 days, how many hours did she work on average per day? At $15.75 per hour, h

Lyrx [107]

She worked 7.5 hours a day on average. You can get this by dividing the 37.5 by 5.

And the average she made was 118.13 when rounded. You can get this by multiplying the 7.5 by the rate

8 0

3 years ago

How do I get X for this question?

exis [7]

Answer:

The value of x is 12

Step-by-step explanation:

To find the value of x, we need to note that the interior angles are equal to 180. We also know that angle R is equal to 180 - (8 + 6x). So we can add all of this together and set equal to 180.

180 - (8 + 6x) + 4x + 2 + 30 = 180

180 - 8 - 6x + 4x + 2 + 30 = 180

-2x + 24 = 0

-2x = -24

x = 12

7 0

3 years ago

It costs $2.80 to make a sandwich at the local deli shop. To make a profit, the deli sells it at a price that is 170% of the cos

Svetlanka [38]

Answer:

Sandwich sells for $4.76 .

Step-by-step explanation:

As given

It costs $2.80 to make a sandwich at the local deli shop.

To make a profit, the deli sells it at a price that is 170% of the cost.

170% is written in the decimal form

$= \frac{170}{100}$

= 1.7

Thus

Sandwich selling cost = 1.7 × Cost of making

Put all the values in the above

Sandwich selling cost = 1.7 × $ 2.80

= $4.76

Therefore the Sandwich sells for $4.76 .

7 0

3 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago