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Paul [167]
2 years ago
11

Patrick climbed 4/5 of the way up the trunk of a tree. Jacob climbed 80/100 of the way up the same tree. To accomplish the same

distance as Patrick and Jacob, how far up that tree trunk will Devon have to climb?
Mathematics
1 answer:
Ainat [17]2 years ago
6 0

Answer:

Standard Description: Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. (Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.)

Step-by-step explanation:

the explanation and answer is there in above

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What is the LCM of 13 and 26
SSSSS [86.1K]
The lest common multiple is 26. If you do multiples of each number

13:13,26
26: 26

Both of their lest common multiples would be 26
5 0
3 years ago
Read 2 more answers
Translate the word phrase into a math expression. nine more than five times a number
levacccp [35]

The number is 'n'.

Five times the number is  5n .

Nine more than that is  5n + 9 .

5 0
3 years ago
SOMEONE PLEASE HELP ME!!! I REALLY NEED SOME HELP!!!
Lostsunrise [7]

Answer:

A. (1, -2)

Step-by-step explanation:

We can substitute the variables of x and y into the inequality of y < -|x|.

Let's start with A, -2 being y and 1 being x.

-2 < - |1|

The absolute value of 1 is 1, and negating that gets us -1.

-2 < -1

Indeed, -2 is less than -1! So A is a solution to the inequality.

Let's test the rest of them, just in case.

For B:

-1 < -|1|

Absolute value of 1 is 1, negating it is -1.

-1

-1 is EQUAL to -1, not less than it, so is not a solution to the inequality.

Let's try C.

0 < -|1|

Absolute value of 1 is 1, negating it is -1.

0 < -1

0 is GREATER than -1, so that is not a solution to the inequality.

Hope this helped!

4 0
3 years ago
What is 4(x-7)=2(x+3)?
hram777 [196]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

4(x−7)=2(x+3)

Simplify both sides of the equation.

4(x−7)=2(x+3)

4x+−28=2x+6

4x−28=2x+6

Subtract 2x from both sides.

4x−28−2x=2x+6−2x

x−28=6

Add 28 to both sides.

2x−28+28=6+28

2x=34

Divide both sides by 2.

2x/2  =  34/2

x = 17

5 0
3 years ago
Read 2 more answers
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
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