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4 years ago

Help plz!!!!!! −3(4 − a) = 6

2 answers:

8 0

_____________________________

Hey!!

Solution,

-3(4-a)=6

or, -12+3a=6

or, 3a=6+12

or, 3a=18

or,a=18/3

a=6

<h3>So the value of X is 6.</h3>

hope it helps..

Good luck on your assignment

inn [45]4 years ago

3 0

Answer:

a = 6

Step-by-step explanation:

use distribution method:

-3(4-a) = 6

-12+3a = 6

3a = 6+12

3a = 18

a = 18/3 = 6

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PLZ HELP! find the value of f(-2) for the function f(x)=4x+10 Show work please

natita [175]

Answer:

f(-2) = +2

Step-by-step explanation:

Let x = -2

So,

f(x) = 4x + 10

Substituting the value of x gives ,

f(-2) = 4 × (-2) + 10 = 10 - 8 = 2

8 0

3 years ago

Jocelyn is 1.75 meters tall. At 1 p.m., she measures the length of a tree's shadow to be

bija089 [108]

The height of the tree = 15.34m

What is height?

Height is a way to measure someone or something from base to top or head to toe. In other words, height tells how tall someone or something is.

Given,

the height of Jocelyn = 1.75m

At 1 pm = 43.85

She is sway from tree = 38.6

therefore, 43.85 - 38.6 = 5.25

Therefore, height

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5.25h = 76.73

therefore, height = 15.34

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8 0

1 year ago

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goldenfox [79]

I think that the answer is A) because the range is y.

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3 years ago

JKLM is a rhombus. KM is 20 and JL is 48. Find the perimeter of the rhombus.

anygoal [31]

Answer:

104 units

Step-by-step explanation:

Given

Shape: Rhombus

$JL = 48$

$KM = 20$

Required

Determine the perimeter

The given parameter are the diagonals of the rhombus.

The perimeter (from diagonals) is calculated as thus:

$P = 2\sqrt{(JL)^2 + (KM)^2}$

Substitute values for JL and KM

$P = 2\sqrt{48^2 + 20^2}$

$P = 2\sqrt{2304 +400}$

$P = 2\sqrt{2704}$

$P = 2 * 52$

$P = 104$

Hence, the perimeter is 104 units

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3 years ago

The two roots a+sqrb and a-sqr b are called _______radicals.

astraxan [27]

Answer:

Conjugate radicals.

Step-by-step explanation:

The two roots a+sqrb and a-sqr b are called conjugate radicals.

6 0

3 years ago