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3 years ago

In a school there are 600 teachers ad 1200 students

Mathematics

1 answer:

8_murik_8 [283]3 years ago

8 0

Answer:

Can you please give a clearer explanation of what is your question?

Step-by-step explanation:

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Which is f(-3) for quadratic function graphed

aniked [119]

At f(-3) the graph is equal to -9.

In order to find the point on the graph, move over to -3 on the x axis. Then move down to find the point where it hits the parabola. You'll see that the parabola hits at point (-3, -9).

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3 years ago

Which power of 2 is equal to this expression: (2•2•2)-2

zhenek [66]

2^3 is equivalent because if you count the number of twos there are in the parentheses its 3.

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3 years ago

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A bag contains 5 red poker chips, 3 blue poker chips and 7 white poker chips. Two chips are drawn without replacement. Find the

prohojiy [21]

5 red, 3 blue, 7 white.....total of 15

probability of 1st drawn being red : 5/15 reduces to 1/3
without replacement
probability of 2nd being red : 4/14 reduces to 2/7

probability of both : 1/3 * 2/7 = 2/21 <==

4 0

3 years ago

Katie has a collection of nickels dimes and quarters with a total value of $4.20 there are 7 more dimes than nickels and six mor

Aneli [31]

Katie, Let the number of nickles = N Since there are 7 more dimes than nickles, the number of dimes D = N + 7 Since there are 4 more quarters than nickles, the number of quarters Q = N + 4 Since quarters = $0.25, dimes = $0.10, and nickles = $0.05 and all added together = $4.90 ⇒ 0.25·Q + 0.10·D + 0.05·N = 4.90 Now multiply through by 100 to eliminate the decimal: 25Q + 10D + 5N = 490 Substitute for Q and D 25(N+4) + 10(N+7) + 5N = 490 25N + 10N + 5N +100 +70 = 490 40N + 170 = 490 ⇒ 40N = 320 ⇒ N = 320/40 N = 8, D = N+7 = 15 Q = N+4 = 12 Check: 25(12) + 10(15) + 5(8) = 300 + 150 + 40 = 490

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