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Angelina_Jolie [31]
3 years ago
7

Plz help me with this

Mathematics
2 answers:
rjkz [21]3 years ago
8 0
12*2+400*2=824
the answer is 824
Afina-wow [57]3 years ago
3 0
824 just add all the sides so 12+12+400+400=824 and in simpler form w2+l2
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PLS HELP !!! Find the missing angle measurements.
dsp73
52=x because x+128=180 and that is a 189 angle
5 0
2 years ago
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Find the missing number: 70, 102, ?, 178, 222
irakobra [83]

Answer:

138

Step-by-step explanation:

70 + 32 = 102

102 + 36 = 138

138 + 40 = 178

178 + 44 = 222

4 0
3 years ago
What is the value of f(-1) when f(x)=2x+2? F(-1)=?
igomit [66]

Here is your function:

f(x)=2x+2

You're looking for f(-1). To get your answer, you have to replace all the x's with -1.

Here is your new equation:

f(-1) = 2(-1)+2

Multiply 2 by -1

2 \times -1 = -2

<em>Remember:</em>

MULTIPLYING

  • Positive(+) times negative(-) is equal to negative
  • Positive(+) times positive(+) is equal to positive
  • Negative(-) times negative(-) is equal to positive
  • Negative(-) times positive(+) is equal to negative

DIVIDING

  • Positive(+) divided by positive(+) is equal to positive
  • Positive(+) divided by negative(-) is equal to negative
  • Negative(-) divided by positive(+) is equal to negative
  • Negative(-) divided by negative(-0 is equal to positive

Since you're multiplying a negative number by a positive number, the answer will be negative.

Here is your new equation:

f(-1) = -2+2

Add -2 and 2:

-2+2 = 0

That means your answer is 0

        ↓

f(-1)=0

<em>If you have any questions, feel free to ask in the comments! :)</em>

5 0
3 years ago
a parallelogram has an area of 288 ft squared. If the base of the parallelogram is 24 ft, what is the height of the parallelogra
labwork [276]

Answer:H=12Ft

Base:24

Area:288

A=bh

h=A/b=288/24=12ft

7 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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