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3 years ago

What is the solution to the system of equations below?

Mathematics

1 answer:

AnnyKZ [126]3 years ago

8 0

Answer:

x=9, y=3

Step-by-step explanation:

-4x+6y=-18

y=-2x+21

-4x+6(-2x+21)=-18

-4x-12x+126=-18

144=16x

x=9

y=-2x+21

y=-18+21

y=3

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If f(x)=e^x-1 +5 and g(x) is a transformation or, the same thing just moved, and its y-intercept is -3 then its equation would be:

D) g(x)=e^x-1 -3

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2 years ago

What is the factor of 16

insens350 [35]

1x16
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3 years ago

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Find the slope of the line that passes through (4,2) and (9,1)?

pshichka [43]

Answer:

The slope of the line is $\displaystyle -\frac{1}{5}$ .

Step-by-step explanation:

We are given two coordinate points:

$(4, 2)$
$(9, 1)$

We are asked to find the slope of the line.

We can use the rise-over-run formula to solve for the slope of the line.

$\displaystyle \text{slope} = \frac{\text{rise}}{\text{run}}\\\\\text{slope} = \frac{y_2-y_1}{x_2-x_1}$

However, we firstly need to name our coordinate points.

In math, we can label our coordinates using the following label system:

$(x_1, y_1), (x_2, y_2)$

Therefore, we can also label our coordinates as such:

$x_1 = 4$
$y_1 = 2$
$x_2 = 9$
$y_2 = 1$

Now, we can supply these values into the formula and solve for our slope, or a better known variable, <em>m</em>.

$\displaystyle m = \frac{1 - 2}{9 - 4}\\\\m = \frac{-1}{5}\\\\m = -\frac{1}{5}$

Therefore, our slope is $\displaystyle -\frac{1}{5}$ .

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2 years ago

Read 2 more answers

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

Part c) How wide are the photo and frame together?

$(4+2x)=4+2(1.5)=7\ in$

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3 years ago

Solve the multi step literal equation: solve equation for A b^2A^2-3g=q

I am Lyosha [343]

Answer:

A = ± $\sqrt{\frac{q+3g}{b^2} }$

Step-by-step explanation:

Given

b²A² - 3g = q ( add 3g to both sides )

b²A² = q + 3g ( divide both sides by b² )

A² = $\frac{q+3g}{b^2}$ ( take the square root of both sides )

A = ± $\sqrt{\frac{q+3g}{b^2} }$

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3 years ago