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Ilia_Sergeevich [38]

3 years ago

9

The histogram below shows the number of sit-ups accomplished by sixth grade students in a physical fitness test. How many studen

ts accomplished 29 sit-ups or less?

1 answer:

Tresset [83]3 years ago

7 0

Can you give a picture of the histogram?

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4 tons of bark cost $29,120.00. What is the price per pound?

son4ous [18]

4*2000

29120/8000=$3.64

7 0

3 years ago

69✖️2[8+12-10✖️2+(7-5)-1]-6*

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69×2 [8+12-10×2+ (7-5) -1] -6 69×2 [8+12-10×2+2-1] -6 69×2 [8+12-20+2-1] -6 69×2 [20-20+2-1] -6 69×2 [0+2-1] -6 69×2 [1] -6] 138 [1] -6 138-6 132 I believe this is ur answer.

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3 years ago

Supplementary angles. If m

Olegator [25]

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Jouc ÷4 i for number 69263

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3 0

2 years ago

Substract the polynomials and simplify (-3r2-2r-4)-(9r2+7r+2)

mina [271]

Answer:

-12r2-9r-6

Step-by-step explanation:

-distribute the subtraction symbol to the second group of numbers

-(9r2+7r+2) = (-9r2-7r-2)

-then combine each term together (-3r2-2r-4)+(-9r2-7r-2)

= -12r2-9r-6

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6 0

3 years ago

Which portion of the unit circle satisfies the trigonometric inequality cos^2theta + sin^2theta is greater than or equal to 1. A

liberstina [14]

Answer:

Only points on the circle satisfy the given inequality.

Step-by-step explanation:

Given: Unit circle

To find: portion of the unit circle which satisfies the trigonometric inequality $\sin ^2\theta +\cos ^2\theta \geq 1$

Solution:

In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)

$\sin \theta$ = side opposite to $\theta$ /hypotenuse

$\cos \theta$ = side adjacent to $\theta$ /hypotenuse

$\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta$

$\cos \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta$

So, coordinates of A = $\left ( \cos \theta ,\sin \theta \right )$

For any point (x,y) on the unit circle with centre at origin, equation of circle is given by $x^2+y^2=1$

Put $(x,y)=\left ( \cos \theta ,\sin \theta \right )$

$\cos ^2\theta +\sin ^2\theta =1$

So, $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ satisfies the equation $x^2+y^2=1$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ inside the circle, $\cos ^2\theta +\sin ^2\theta$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ outside the circle, $\cos ^2\theta +\sin ^2\theta >1$

So, only points on the circle satisfy the given inequality.

4 0

2 years ago

Read 2 more answers