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Katyanochek1 [597]
3 years ago
7

WHOEVER ANSWERS CORRECT I'LL MARK BRAINLIEST

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

The missing value is 21.

The ratio value of the double number line is \frac{7}{2}.

Solution:

Ratio of the given number line is

$\frac{7}{2}=\frac{7\div1}{2\div1}=\frac{7}{2}

$\frac{14}{4}=\frac{14\div2}{4\div2}=\frac{7}{2}

$\frac{28}{8}=\frac{28\div4}{8\div4}=\frac{7}{2}

$\frac{35}{10}=\frac{35\div5}{8\div5}=\frac{7}{2}

Hence the common ratio of the number line is \frac{7}{2}.

The missing ratio is

$\frac{7}{2} =\frac{7\times3}{2\times3}= \frac{21}{6}

Hence the missing value is 21.

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Bathing suits are on sale 80% off at Dillards. If I see a bathing suit marked down to $25, what was the original price of the ba
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there are 55 students in a band 11 are 6th graders 21 are 7th grader and 23 are 8th graders what is the ratio of the 6th graders
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11 : 44

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3 years ago
Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation
just olya [345]

Answer:

The particle will travel 6 feet in first 2 seconds.

Step-by-step explanation:

We have been given that a particle moves according to the velocity equation v(t)= 6t^2-18t+12. We are asked to find the distance that the particle will travel in its first 2 seconds.

s(t)=\int |v(t)|dt

s(t)=\int\limits^2_0 |6t^2-18t+12|dt

Now, we will eliminate the absolute value sign as:

s(t)=\int\limits^1_0 6t^2-18t+12dt+\int\limits^2_1 -6t^2+18t-12dt

s(t)=[\frac{6t^3}{3}-\frac{18t^2}{2}+12t]^1_0 +[\frac{-6t^3}{3}+\frac{18t^2}{2}-12t]^2_1

s(t)=[2t^3-9t^2+12t]^1_0 +[-2t^3+9t^2-12t]^2_1

s(2)=2(1)^3-9(1)^2+12(1)-(2(0)^3-9(0)^2+12(0))-2(2)^3+9(2)^2-12(2)-(-2(1)^3+9(1)^2-12(1))

s(2)=2-9+12-(0)-16+36-24-(-2+9-12)

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s(2)=5-4+5

s(2)=6

Therefore, the particle will travel 6 feet in first 2 seconds.

   

 

7 0
3 years ago
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