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3 years ago

5

Solve the following system of equations. Express your answer as an ordered pair in the format (a,b), with no spaces between the

numbers or symbols. First equation is 3x+4y=17 And Second equation is -4x-3y=-18

2 answers:

Mumz [18]3 years ago

6 0

I have no idea maybe ask ur math teacher

expeople1 [14]3 years ago

6 0

Answer:

(3, 2)

Step-by-step explanation:

The system of equations can be solved using elimination to eliminate one variable and solve for the other. Given the two equations:

3x + 4y = 17 and -4x - 3y = -18, we need to have opposite coefficients to eliminate a variable. Multiplying the first equation by '3' and the second by '4':

3(3x + 4y = 17) = 9x + 12y = 51

4(-4x - 3y = -18) = <u>-16x - 12y = -72</u>

-7x = -21

x = 3

Solve for y: 3(3) + 4y = 17 or 9 + 4y = 17

y = 2

(3, 2)

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What is the solution -10 = 2a - 4

Simora [160]

Add 4 to both sides

-10 + 4 = 2a

Simplify -10 + 4 to -6

-6 = 2a

Divide both sides by 2

-6/2 = a

Simplify 6/2 to 3

-3 = a

Switch sides

<u>a = -3</u>

6 0

3 years ago

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Write an equation of a parabola that passes through (3,-30) and has x-intercepts of -2 and 18. Then find the average rate of cha

Nookie1986 [14]

Answer:

The equation of the parabola is $y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}$ . The average rate of change of the parabola is -4.

Step-by-step explanation:

We must remember that a parabola is represented by a quadratic function, which can be formed by knowing three different points. A quadratic function is standard form is represented by:

$y = a\cdot x^{2}+b\cdot x + c$

Where:

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

$a$ , $b$ , $c$ - Coefficients, dimensionless.

If we know that $(3, -30)$ , $(-2, 0)$ and $(18, 0)$ are part of the parabola, the following linear system of equations is formed:

$9\cdot a +3\cdot b + c = -30$

$4\cdot a -2\cdot b +c = 0$

$324\cdot a +18\cdot b + c = 0$

This system can be solved both by algebraic means (substitution, elimination, equalization, determinant) and by numerical methods. The solution of the linear system is:

$a = \frac{2}{5}$ , $b = -\frac{32}{5}$ , $c = -\frac{72}{5}$ .

The equation of the parabola is $y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}$ .

Now, we calculate the average rate of change ( $r$ ), dimensionless, between $x = -2$ and $x = 8$ by using the formula of secant line slope:

$r = \frac{y(8)-y(-2)}{8-(-2)}$

$r = \frac{y(8)-y(-2)}{10}$

$x = -2$

$y = \frac{2}{5}\cdot (-2)^{2}-\frac{32}{5}\cdot (-2)-\frac{72}{5}$

$y(-2) = 0$

$x = 8$

$y = \frac{2}{5}\cdot (8)^{2}-\frac{32}{5}\cdot (8)-\frac{72}{5}$

$y(8) = -40$

$r = \frac{-40-0}{10}$

$r = -4$

The average rate of change of the parabola is -4.

3 0

3 years ago

I need help on this answer and don't know. Do you know how to do it?????

tatyana61 [14]

0 + (-3 1/4)
1 + (-4 1/4)
2 + (-5 1/4)

7 0

3 years ago

42 pieces of candy divided between three people. You put candy in three boxes. How many pieces of candy are there in per box?

Salsk061 [2.6K]

Answer:

There are 14 pieces of candy in each box.

Step-by-step explanation:

42 divided by 3 equals 14.

<u><em>Could I please have Brainliest.</em></u>

5 0

3 years ago

A restaurant menu has five kinds of soups, seven kinds of main courses, six kinds of desserts, and five kinds of drinks. If a cu

ryzh [129]

Answer: 1050

Step-by-step explanation:

Number of combinations of selecting r things out of n = $^nC_r=\dfrac{n!}{r!(n-r)!}$

such that $^nC_1=n$

Given: A restaurant menu has 5 kinds of soups, 7kinds of main courses, 6 kinds of desserts, and 5 kinds of drinks.

If a customer randomly selects one item from each of these four categories, then by fundamental counting principle , the number of different outcomes are possible = $^5C_1\times \ ^7C_1\times\ ^6C_1\times\ ^5C_1 =5\times7\times6\times5=1050$

hence, total number of outcomes = 1050

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3 years ago

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