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3 years ago

Simplifying algebraic expressions 4t+3t-5-7t

Mathematics

1 answer:

Elden [556K]3 years ago

5 0

4t + 3t - 5 - 7t
answer: -5

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3 years ago

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Solve 7r+ 2 = 5(r – 4)

AveGali [126]

Multiply the bracket by 5

I used PEMDAS

P= parenthesis

E= exponents

M=multiplication

D= division

A= addition

S= subtraction

7r+2= 5(r-4)

7r+2= 5r-20

Move 5r to the left hand side . Positive 5r changes to negative 5r

7r-5r+2= 5r-5r-20

2r+2=- 20

2r+2-2= -20-2

Move positive 2 to the right hand side. Changes to negative -2

2r+2-2= -20-2

2r= -22

Divide by 2 for 2r and -22

2r/2= -22/2

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Answer is r= -11

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3 years ago

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Geometry Help! 30 POINTS! + BRAINLIEST

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SRT = 20 => STR = 20.
Since STR and STU are supplementary angles, STU = 180 - STR = 180 - 20 = 160, means x = 40.

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4 years ago

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Digiron [165]

Extraneous root since dividing by zero is undefined

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3 years ago

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

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3 years ago

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