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3 years ago

the top of a utility pole is 14 feet high. The base of the pole is at -2 feet. Find the length of the utility pole.

Mathematics

2 answers:

Nutka1998 [239]3 years ago

8 0

16 feet, you change the -2 to 2 and add. Hope this helps.

lina2011 [118]3 years ago

6 0

16 will be your answer

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Consider a $2 principal investment with a 10% annual simple interest rate. Enter the simple interest equation that represents th

PtichkaEL [24]

The simple interest equation that represent the situation is A = t/5

<h3>What is simple interest?</h3>

Simple Interest (S.I.) is the method of calculating the interest amount for a particular principal amount of money at some rate of interest. For example, when a person takes a loan of 500 dollars, at a rate of 10 per annum for two years, the person's interest for two years will be S.I. on the borrowed money.

Principal(P) borrowed = $2

Rate(R) in percentage = 10

time in years = t

Simple interest(A) = P X R X t / 100

A = 2 X 10 X t/ 100

After reducing 20/100 to the lowest term

A = t/5

In conclusion the simple interest(A) = t/5

Learn more about simple interest: brainly.com/question/20690803

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5 0

1 year ago

me. Evans needs to assign 32 students into 8 equal groups. he says "I can use repeated subtraction. since I subtract 3 times, ea

natali 33 [55]

Hi there.

So, Mr. Evans wants to divide 32 students into 8 equal groups. While he wants to use repeated subtraction of 3, the correct answer is to use division of 8.

Let's put that into math.

32/8 = 4

I don't agree with Mr. Evans because 32 students divided equally into groups of 8 means 4 students per group.

Let me know if this was easy to understand :)

4 0

3 years ago

What are the factors of a product of -33 and a sum of 8

mars1129 [50]

Answer:

-3 and 11

Step-by-step explanation:

I basically drew a t-chart and I knew that -33 is the product (mutiplies to) and 8 is what it adds to. The only numbers that work are -3 and 11 because it says positive 8. -3 and 11 add to +8. Hope this helps!

7 0

3 years ago

Hey I’m confused can you help?

Kamila [148]

The interquartile range (IQR) is calculated by subtracting the first quartile from the third quartile

so Q3 - Q1

answer: A

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%5Cfrac%7Bd%7D%7Bdx%7D%20%20%20%5Cleft%20%28%20%5Cbigg%28%20%5Cint_%7B1%7D%5E%7B%20%7B

Margaret [11]

Applying the product rule gives

$\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}$

Use the fundamental theorem of calculus to compute the remaining derivatives.

$\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt$

The remaining integrals are

$\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}$

$\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)$

and so the overall derivative is

$\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)$

which could be simplified further.

5 0

2 years ago