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2 years ago

Solve the following system of equations. What is one x-value of a solution? f(x)=x^(2)+7x-24 g(x)=-3x-40

Mathematics

1 answer:

Yuki888 [10]2 years ago

5 0

One x- value of the solution is : -2 or -8

<h3>What is an equation?</h3>

An Equation is an algebraic statement that shows that two quantities are equal.

Analysis:

for one x-value of a solution, f(x) = g(x)

$x^{2}$ + 7x -24 = -3x - 40

$x^{2}$ +7x +3x -24 +40 = 0

$x^{2}$ + 10x +16 = 0

$x^{2}$ +8x +2x +16 = 0

x(x+8) +2(x+8) = 0

(x+2)(x+8) = 0

x = -2 or -8

In conclusion, the values of x are -2 or -8

Learn more about system of equation : brainly.com/question/14323743

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3 years ago

5. A researcher creates a scatter plot that displays the relationship between the number of years in business, , and the percent

Black_prince [1.1K]

Based on the information given, it should be noted that the residual for the two points will be 0.087 and -0.033 respectively.

<h3>How to find the residual.</h3>

From the complete information, the predicted value for the point (3, 0.42) will be:

= (0.091 × 3) + 0.060

= 0.0273 + 0.060

= 0.333

Therefore, the residual will be:

= 0.42 - 0.333 = 0.087

The predicted value for the point (3, 0.3) will be:

y = 0.091x + 0.060.

= (0.091 × 3) + 0.060.

= 0.333

Therefore, the residual will be:

= 0.3 - 0.333 = -0.033

Therefore, the residual for the two points will be 0.087 and -0.333 respectively.

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2 years ago

A spherical- shaped ball that has a radius of 4 inches is currently deflated.

galina1969 [7]

Answer:

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3 years ago

What is y=4/5x+7 in standard form?

viktelen [127]

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3 years ago

P(x)=Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,12).

Mila [183]

Answer:

The polynomial is:

$p(x) = -x^3 - 2x^2 + 5x + 6$

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots $x_{1}, x_{2}, x_{n}$ such that it can be written as: $a(x - x_{1})*(x - x_{2})*...*(x-x_n)$ , in which a is the leading coefficient.

Zeros of −3, −1, and 2

This means that $x_1 = -3, x_2 = -1, x_3 = 2$ . Thus

$p(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)$

$p(x) = a(x - (-3))*(x - (-1))*(x-2)$

$p(x) = a(x+3)(x+1)(x-2)$

$p(x) = a(x^2+4x+3)(x-2)$

$p(x) = a(x^3+2x^2-5x-6)$

Passes through the point (1,12).

This means that when $x = 1, p(x) = 12$ . We use this to find a.

$12 = a(1 + 2 - 5 - 6)$

$-12a = 12$

$a = -\frac{12}{12}$

$a = -1$

Thus

$p(x) = -(x^3+2x^2-5x-6)$

$p(x) = -x^3 - 2x^2 + 5x + 6$

6 0

3 years ago