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Mkey [24]
3 years ago
11

the side of the clip board appears to be a right triangle . the leg length are 2 millimeters and 2.1 millimeters and the hypoten

use is 2.9 millimeters. is the thee of the clip a right triangle ?
Mathematics
1 answer:
givi [52]3 years ago
5 0

It is because u can only do pathagreom therom on right triangles.

2*2 + 2.1*2.1

4+4.41=8.41

square root of 8.41 is 2.9 so that confrms the therom which isaversa confirms it is a right triangle.

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What is the equation of the line expressed in slope-intercept form?
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3 years ago
Solve the absolute value inequality |x+9| > 3
Studentka2010 [4]

Answer:

x>−6,x<−12

Step-by-step explanation:

Break down the problem into these 2 equations.

x+9>3x+9>3

-(x+9)>3−(x+9)>3

Solve the 1st equation: x+9>3x+9>3.

x>-6x>−6

3 Solve the 2nd equation: -(x+9)>3−(x+9)>3.

x<-12x<−12

Collect all solutions.

x>-6,x<-12x>−6,x<−12

8 0
3 years ago
I WILL GIVE BRAINLIEST!!!
netineya [11]

Answer:

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

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Step-by-step explanation:i really hope this helps you

7 0
3 years ago
A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)
DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

Speed=\frac{distance}{time}

in this problem we have

distance=400\ m \\time=120\ sec

substitute in the formula

Speed=\frac{400}{120}

Speed=3.5\frac{m}{sec}

therefore

<u>the answer Part a) is</u>

the speed is equal to 3.5\frac{m}{sec}

<u>Part b) </u>Find the velocity

we know that

<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

magnitude=3.5\frac{m}{sec}

direction=North\ East\ (NE)

therefore

<u>the answer Part b) is</u>

the velocity is 3.5\frac{m}{sec}\ North\ East\ (NE)

Part c)

we know that

the acceleration is equal to the formula

a=\frac{V2-V1}{t2-t1}

in this problem we have

V2=0 \\V1=3.5\frac{m}{sec}

t2=15\ sec\\t1=0

substitute in the formula

a=\frac{0-3.5}{15-0}

a=-\frac{3.5}{15}\frac{m}{sec^{2}}

a=-\frac{7}{30}\frac{m}{sec^{2}}

therefore

<u>the answer Part c) is</u>

the acceleration is -\frac{7}{30}\frac{m}{sec^{2}}

This is an example of negative acceleration

7 0
3 years ago
Read 2 more answers
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