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Arisa [49]
3 years ago
12

Which is a graph for the inequality m < -2 ?

Mathematics
1 answer:
aliya0001 [1]3 years ago
6 0

Answer:

B

Step-by-step explanation:

This answer shows values of -2, and less than -2

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a giant pizza box in the shape of a rectangular prism measures 38m times 38m times 0.25m. what is the surface area of the box
Helga [31]
So the surface area of a box has this equation: A = 2 (wl + hl + hw)

w = width = 38
l = length = 38
h = height = 0.25

Plugging in:
A = 2 ((38*38) + (38*0.25) + (38*0.25))
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3 years ago
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Solve for v simplify your answer as much as possible: -2 + 3v = -23​
seropon [69]

Answer:

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Step-by-step explanation:

-2 + 3v = -23 >> -2 + 23 = -3v >> 21 = -3v >> v = -7

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2 years ago
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Which algebraic expression represents "the difference of 54 and a number"?
atroni [7]

\huge{\boxed{54-x}}

The difference is the result of a subtraction problem.

We are given two values that are being subtracted: 54 and a number, represented by x

So, represent this mathematically with \boxed{54-x}.

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3 years ago
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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

8 0
3 years ago
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PLSSSS HELP ME WITH THIS QUESTION!!!
AysviL [449]

Given,  

∠AOE = 30°

∠DOB = 40°

So, ∠EOD = 180 - (30+40)           [angles on a straight line]

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∠EOD = ∠COF                             [ vertically opposite angle]

∴ ∠COF = 110°

7 0
3 years ago
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