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alukav5142 [94]

4 years ago

5

Plz help I will mark you brainlist plz

1 answer:

emmasim [6.3K]4 years ago

3 0

imma attach a screenshot of the work and answers to those equations

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Helaine graphed the equation 12x−4y=3. What was the slope of Helaine’s line?

zubka84 [21]

Answer:

3

X is slope but the equation isnt set up proper

12x-4y=3

-12x -12x

-4y=-12x+3

/-4 /-4

y= 3x+3/-4

8 0

3 years ago

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Help please <br> -x^3 = 216<br> a. +/- 6<br> B. 6<br> C. -6 ∛6<br> D. -6

rosijanka [135]

Answer:

D. -6

Step-by-step explanation:

SInce x is negative, the - before the six would be cancelled out leaving the equation to be 6^3=216 which is correct

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3 years ago

By rounding each number to 1 significant figure,

luda_lava [24]

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Step-by-step explanation:

1. 8.2 * 6.7 / 0.46

2. 54.94 / 0.46

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3 years ago

Which of the following equations is FALSE?<br>QT = 16<br> RS = 32 <br>QS= 32 <br>RT = 16

DanielleElmas [232]

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7 0

2 years ago

PLease answer !!! Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all

Xelga [282]

Answer:

1. $(A,B) = (3,-2)$

2. The values of t are: -3, -1

Step-by-step explanation:

Given

$\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$

$|t| = 2t + 3$

Required

Solve for the unknown

Solving $\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$

Take LCM

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{(x - 2)(x-1)}$

Expand the denominator

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - 2x + x -2}$

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - x -2}$

Both denominators are equal; So, they can cancel out

$x + 7 = A(x+1) + B(x-2)$

Expand the expression on the right hand side

$x + 7 = Ax + A + Bx - 2B$

Collect and Group Like Terms

$x + 7 = (Ax + Bx) + (A - 2B)$

$x + 7 = (A + B)x + (A - 2B)$

By Direct comparison of the left hand side with the right hand side

$(A + B)x = x$

$A - 2B = 7$

Divide both sides by x in $(A + B)x = x$

$A + B = 1$

Make A the subject of formula

$A = 1 - B$

Substitute 1 - B for A in $A - 2B = 7$

$1 - B - 2B = 7$

$1 - 3B = 7$

Subtract 1 from both sides

$1 - 1 - 3B = 7 - 1$

$-3B = 6$

Divide both sides by -3

$B = -2$

Substitute -2 for B in $A = 1 - B$

$A = 1 - (-2)$

$A = 1 + 2$

$A = 3$

Hence;

$(A,B) = (3,-2)$

Solving $|t| = 2t + 3$

Because we're dealing with an absolute function; the possible expressions that can be derived from the above expression are;

$t = 2t + 3$ and $-t = 2t + 3$

Solving $t = 2t + 3$

Make t the subject of formula

$t - 2t = 3$

$-t = 3$

Multiply both sides by -1

$t = -3$

Solving $-t = 2t + 3$

Make t the subject of formula

$-t - 2t = 3$

$-3t = 3$

Divide both sides by -3

$t = -1$

<em>Hence, the values of t are: -3, -1</em>

7 0

3 years ago

Read 2 more answers