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3 years ago

HELPP PLEASE!!!!!! ):

Mathematics

1 answer:

padilas [110]3 years ago

8 0

Answer:

-3
$\frac{3}{4}$
$\frac{-1}{3}$
$\frac{3}{16}$
$\frac{3}{10000}$

Step-by-step explanation:

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-3(1-2x)-3=-20-8x what is the work and answer

larisa86 [58]

You have to keep order of operations in mind.

1. Take care of the parentheses; distribute.
-3+6x-3=-20-8x
Simplified: 6x-6=-20-8x

2. Isolate the x variable.
Add 8x to both sides and add 6 to both sides.
14x=-14

3. Solve for x.
Divide by 14 on both sides.
x=-1

4 0

3 years ago

Please help asap. 45 points

Levart [38]

The answer is B. hope that help

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3 years ago

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A rectangle has an area of 40 square units. The length is 6<br> units greater than the width.

stiks02 [169]

Answer:

Answer:

The width is 4 units, and the length is 10 units.

Step-by-step.

Step-by-step explanation:

area of rectangle = length * width

Let L = length; let W = width.

"The length is 6 units greater than the width.": L = W + 6

area = LW = 40

Since L = W + 6, we substitute L with W + 6.

(W + 6)W = 40

W^2 + 6W = 40

W^2 + 6W - 40 = 0

(W - 4)(W + 10) = 0

W - 4 = 0 or W + 10 = 0

W = 4 or W = -10

A width cannot be a negative number, so we discard the solution W = -10.

W = 4

L = W + 6 = 4 + 6 = 10

The width is 4 units, and the length is 10 units.

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3 years ago

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hannah made 18 ounces of applesauce. She evenly portions the applesauce among 4 containers. how many ounces of applesauce are in

Fittoniya [83]

4.5 ounces of applesauce in 1 container.

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3 years ago

Solve with cramer's rule x+2y+3z=11, 2x+y+2z=10, 3x+2y+z=9

nalin [4]

Answer:

x = 2 , y = 0 , z = 3

Step-by-step explanation:

Cramer's rule is a rule through which we can find the solution of linear equation.

we have the three linear equations as

x+2y+3z=11

2x+y+2z=10

3x+2y+z=9

AX=B

A: coefficient matrix

X= unknown vectors(x,y,z)

D = values of the linear equation (11 , 10 , 9)

now we find the determinant of the given linear equation

determinant of the matrix will be

A = $\left[\begin{array}{ccc}1&2&3\\2&1&2\\3&2&1\end{array}\right]$ = 1(1-4) - 2(2-6) + 3(4 - 3)

= 1(-3) - 2(-4) + 3(1)

= -3+8+3 = 8

also D $\neq 0$

so the determinant is Non zero we can apply Cramer's rule

we will be replacing the first column of the coefficient matrix A with the values of D

by replacing the first column we will get the value of the variable 'x'

Dx = $\left[\begin{array}{ccc}11&2&3\\10&1&2\\9&2&1\end{array}\right]$ = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16

x = $\frac{Dx}{D}$ = $\frac{16}{8}$ = 2

similarly

Dy = $\left[\begin{array}{ccc}1&11&3\\2&10&2\\3&9&1\end{array}\right]$ = 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0

y = $\frac{Dy}{D}$ = 0

Dz= $\left[\begin{array}{ccc}1&2&11\\2&1&10\\3&2&9\end{array}\right]$ = 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24

z = $\frac{Dz}{D}$ = $\frac{24}{8} = 3$

so we have the solution as

x = 2 , y = 0 , z = 3

Therefore the solution for the given linear equations is (2,0,3).

3 0

3 years ago