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3 years ago

A packaging employee making $16

Mathematics

1 answer:

Vaselesa [24]3 years ago

7 0

Answer:

Step-by-step explanation:

that's the amount is t coe

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With show work please. urgent

son4ous [18]

Answer:

answer is 3 or 50 degrees

Step-by-step explanation:

put angle 35 degrees between F and E and then add 5 degrees to get 40

then add the 90 degrees to get 130

then subtract 130 from 180 to get 50 degress

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2 years ago

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6. Find the slope of the line that passes through the points: (4, 5) and (-2, 8) Vour answer

Marina CMI [18]

Answer:

m=-1/2

Step-by-step explanation:

m=y2-y1/x2-x1

m=8-5/-2-4

m=3/-6

m=-1/2

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3 years ago

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HELP, I WILL GIVE BRAINLIEST!!!!! (ignore my writing)

sveticcg [70]

Sorry if some things aren’t rounded correctly :)

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2 years ago

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Y equals equation for the line that passes through the points (-6,2) and (1,16)

8_murik_8 [283]

Slope-intercept form: y = mx + b

[m is the slope, b is the y-intercept or the y value when x = 0 ---> (0, y) or the point where the line crosses through the y-axis]

To find the slope, you use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$ And plug in the two points

(-6, 2) = (x₁ , y₁)

(1, 16) = (x₂ , y₂)

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{16-2}{1-(-6)}$ [two negatives cancel each other out and become positive]

$m=\frac{16-2}{1+6}$

$m=\frac{14}{7}$

m = 2 Now that you found m, plug it into the equation

y = mx + b

y = 2x + b To find b, plug in one of the points into the equation

(-6, 2)

y = 2x + b

2 = 2(-6) + b

2 = -12 + b Add 12 on both sides

14 = b

(1, 16)

y = 2x + b

16 = 2(1) + b Subtract 2 on both sides

14 = b

Now that you found b, plug it into the equation

y = 2x + 14

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3 years ago

ANSWER ASAP PLEASE & SHOW UR WORK!!!

Lina20 [59]

Answer:

$x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}$

Step-by-step explanation:

<u>Biquadratic Equations</u>

Solve:

$x^4 + 3x^2 - 18 = 0$

The biquadratic equations are equations of degree 4 without the terms of degree 1 and 3.

Solving such equations requires to express the equation as a second-degree equation with $x^2$ as the variable.

Rewriting the equation:

$(x^2)^2 + 3(x^2) - 18 = 0$

The quadratic equation can be factored as:

$(x^2+6)(x^2-3)=0$

It leads to two equations:

$x^2+6=0$

$x^2-3=0$

The first equation has imaginary roots. Solving for x:

$x^2=-6$

$x=\pm\sqrt{-6}$

$x_1=\mathbf{i}\sqrt{6}$

$x_2=-\mathbf{i}\sqrt{6}$

Where

$\mathbf{i}=\sqrt{-1}$

The second equation has two real roots:

$x^2-3=0$

$x^2=3$

$x=\pm\sqrt{3}$

$x_3=\sqrt{3}$

$x_4=-\sqrt{3}$

The roots are:

$\mathbf{x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}}$

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3 years ago