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2 years ago

A 32 foot tall tree casts a shadow that is 48 feet long. How far away from the tree is a 6 ft person? Help! :)

Mathematics

1 answer:

shusha [124]2 years ago

4 0

Answer:

Henry Clay believed that American democracy depended on “that great principle of compromise”. What did he mean? Do you agree? Why?12

Step-by-step explanation:

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A team of runners is needed to run a 1/3-mile relay race. If each runner must run 1/9 mile, how many runners will be needed?

Alex Ar [27]

Answer:

3 Runners

Step-by-step explanation:

6 0

2 years ago

How many 1/3 cups of ice cream can you make from 1/7 cup

OlgaM077 [116]

Well first you have to make the denominators equal by finding a common denominator. in this case 21, but whatever you do to the bottom you do to the top. so 1/3 times 7/7 equals 7/21 and 1/7 times 3/3 equals 3/21.

so how many 7/21 cups of ice cream can you make from 3/21 i think about none but remember i could be wrong

8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1.107%7D%7B123%7D%20%20%5Ctimes%20%20%5Cfrac%7B1.998%7D%7B333%7D%20%20%3D%20" id=

pav-90 [236]

Answer:

Step-by-step explanation:

$\frac{1.107}{123} \times \frac{1.998}{333} =$

$= > 9 \times 6$

$= 45$

6 0

2 years ago

Read 2 more answers

Construct and interpret a 95% confidence interval estimate for the difference in the mean number of cell phone calls per month f

dsp73

Answer:

25 per month

Step-by-step explanation:

5 0

3 years ago

Write the equation of a polynomial of degree 3, with zeros 1, 2 and -1 where f(0)=2

drek231 [11]

<u>Answer:</u>

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

<u>Solution:</u>

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

$F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0$

where a, b, c are roots of the polynomial.

Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$

So, the equation is $x^{3}-2 x^{2}-x+2=0$

Let us put x = 0 in f(x) to check whether our answer is correct or not.

$\mathrm{F}(0) \rightarrow 0^{3}-2(0)^{2}-0+2=2$

Hence, the equation of the polynomial is $x^{3}-2 x^{2}-x+2=0$

3 0

2 years ago