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kifflom [539]
3 years ago
12

June bought ​ 3 1 2 312 ​ gallons of white paint at $17.50 per gallon and ​ 1 3 4 134 ​gallons of blue paint at $19.00 per gallo

n. What was her total cost?
Mathematics
1 answer:
algol [13]3 years ago
5 0
312* 17.5= 5460.00

134*19=2546

2546+5460= $8,006.00
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Find the principal (positive) square root of 4. If necessary, round the answer to the nearest hundredth (two decimal places).
Lana71 [14]
The principal square root of 4 is 2
8 0
3 years ago
(b) The area of a rectangular painting is 7719 cm2.
Nataliya [291]

Given:

  • The area of a rectangular painting is 7719 cm²
  • Width of painting = 83cm

To Find?

  • Length of painting.

Solution:

  • Let length of painting be x cm

Using formula:

  • Area of rectangle = L × B

Where,

  • Length = x cm
  • Breadth = 73 cm

→ 7719 = x × 83

→ x = 7719/83

→ x = 93 cm

Hence,

  • Length of painting is 93cm
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2 years ago
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Is 40/25 equivalent to 32/20
Artyom0805 [142]

Answer:

Yes, 40/25 is equivalent to 32/20.

8 0
3 years ago
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What are the solutions of the system?
miskamm [114]
We have the following system of equations:
 y = -6x-6
 y = x2-5x-6
 Matching we have:
 x2-5x-6 = -6x-6
 Rewriting we have:
 x2-5x-6 + 6x + 6 = 0
 x2 + x = 0
 Rewriting:
 x (x + 1) = 0
 The solutions are:
 x = 0
 x = -1
 For x = 0
 y = -6 (0) -6 = 0-6 = -6
 y = -6
 For x = -1
 y = -6 (-1) -6 = 6-6 = 0
 y = 0
 The solutions are:
 (x, y) = (0, -6)
 (x, y) = (- 1, 0)
 Answer:
 
The solutions of the system are:
 (-1, 0) and (0, -6)
3 0
3 years ago
Solve using elimination<br> x+y-2z=8<br> 5x-3y+z=-6<br> -2x-y+4z=-13
Free_Kalibri [48]
So here is your answer with LaTeX issued format interpretation. Full process elucidated briefly, below:

\begin{alignedat}{3}x + y - 2z = 8 \\ 5x - 3y + 2 = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}

For this equation to get obtained under the impression of those variables we have to eliminate them individually for moving further and simplifying the linear equation with three variables along the axis.

Multiply the equation of x + y - 2z = 8 by a number with a value of 5; Here this becomes; 5x + 5y - 10z = 40; So:

\begin{alignedat}{3}5x + 5y - 10z = 40 \\ 5x - 3y + z = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}

Pair up the equations in a way to eliminate the provided variable on our side, that is; "x":

5x - 3y + z = - 6

-

5x + 5y - 10z = 40
______________

- 8y + 11z = - 46

Therefore, we are getting.

\begin{alignedat}{3}5x + 5y - 10z = 40 \\ - 8y + 11z = - 46 \\ - 2x - y + 4z = - 13 \end{alignedat}

Multiply the equation of 5x + 5y - 10z = - 40 by a number with a value of 2; Here this becomes; 10x + 10y - 20z = 80.

Multiply the equation of - 2x - y + 4z = - 13 by a number with a value of 5; Here this becomes; - 10x - 5y + 20z = - 65; So:

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ - 10x - 5y + 20z = - 65 \end{alignedat}

Pair up the equations in a way to eliminate the provided variables on our side, that is; "x" and "z":

- 10x - 5y + 20z = - 65

+
10x + 10y - 20z = 80
__________________

5y = 15

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ 5y = 15 \end{alignedat}

Multiply the equation of - 8y + 11z = - 46 by a number with a value of 5; Here this becomes; - 40y + 55z = - 230.

Multiply the equation of 5y = 15 by a number with a value of 8; Here this becomes; 40y = 120; So:

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 690 \\ 40y = 120 \end{alignedat}

Pair up the equations in a way to eliminate the provided variables on our side, that is; "y":

40y = 120

+

- 40y + 55z = - 230
_________________

55z = - 110

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 230 \\ 55z = - 110 \end{alignedat}

Solving for the variable of 'z':

\mathsf{55z = - 110}

\bf{\dfrac{55z}{55} = \dfrac{-110}{55}}

Cancel out the common factor acquired on the numerator and denominator, that is, "55":

z = - \dfrac{\overbrace{\sout{110}}^{2}}{\underbrace{\sout{55}}_{1}}

\boxed{\mathbf{z = - 2}}

Solving for variable "y":

\mathbf{\therefore \quad - 40y - 55 \big(- 2 \big) = - 230}

\mathbf{- 40y - 55 \times 2 = - 230}

\mathbf{- 40y - 110 = - 230}

\mathbf{- 40y - 110 + 110 = - 230 + 110}

Adding the numbered value as 110 into this equation (in previous step).

\mathbf{- 40y = - 120}

Divide by - 40.

\mathbf{\dfrac{- 40y}{- 40} = \dfrac{- 120}{- 40}}

\mathbf{y = \dfrac{- 120}{- 40}}

\boxed{\mathbf{y = 3}}

Solve for variable "x":

\mathbf{10x + 10y - 20z = 80}

\mathbf{Since, \: z = - 2; \quad y = 3}

\mathbf{10x + 10 \times 3 - 20 \times (- 2) = 80}

\mathbf{10x + 10 \times 3 + 20 \times 2 = 80}

\mathbf{10x + 30 + 20 \times 2 = 80}

\mathbf{10x + 30 + 40 = 80}

\mathbf{10x + 70 = 80}

\mathbf{10x + 70 - 70 = 80 - 70}

\mathbf{10x = 10}

Divide by this numbered value \mathbf{10} to get the final value for the variable "x".

\mathbf{\dfrac{10x}{10} = \dfrac{10}{10}}

The numbered values in the numerator and the denominator are the same, on both the sides. This will mean the "x" variable will be left on the left hand side and numbered values "10" will give a product of "1" after the division is done. On the right hand side the numbered values get divided to obtain the final solution for final system of equation for variable "x" as "1".

\boxed{\mathbf{x = 1}}

Final solutions for the respective variables in the form of " (x, y, z) " is:

\boxed{\mathbf{\underline{\Bigg(1, \: \: 3, \: \: - 2 \Bigg)}}}

Hope it helps.
8 0
3 years ago
Read 2 more answers
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