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Whitepunk [10]
2 years ago
12

URGENT HELP ME OUT PLEASE 20 points What are the solutions to the quadratic equation?

Mathematics
1 answer:
icang [17]2 years ago
8 0
I think it’s c hope this helps you :)
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party monkey charges $75 for set up and $8 per person. birthday central charges $50 for set-up and $11 per person. for how many
qaws [65]

Answer:

9 or more

Step-by-step explanation:

5 0
3 years ago
Solve 3x+2y=5 and 7x+2y=9
vekshin1

Answer:

Add the equations in order to solve for the first variable. Plug this value into the other equations in order to solve for the remaining variables.

Point Form:(1,1)

Step-by-step explanation:

6 0
2 years ago
A shooting star forms a right triangle with the Earth and the Sun, as shown below:
Semmy [17]

I found this!!!!

The scientist can use these two measurements to calculate the distance between the Sun and the shooting star by applying one of the trigonometric functions: Cosine of an angle.

- The scientist can substitute these measurements into cos\alpha=\frac{adjacent}{hypotenuse}cosα=

hypotenuse

adjacent

and solve for the distance between the Sun and the shooting star (which would be the hypotenuse of the righ triangle).

Step-by-step explanation:

You can observe in the figure attached that "AC" is the distance between the Sun and the shooting star.

Knowing the distance between the Earth and the Sun "y" and the angle x°, the scientist can use only these two measurements to calculate the distance between the Sun and the shooting star by applying one of the trigonometric functions: Cosine of an angle.

This is:

cos\alpha=\frac{adjacent}{hypotenuse}cosα=

hypotenuse

adjacent

In this case:

\begin{gathered}\alpha=x\°\\\\adjacent=BC=y\\\\hypotenuse=AC\end{gathered}

α=x\°

adjacent=BC=y

hypotenuse=AC

Therefore, the scientist can substitute these measurements into cos\alpha=\frac{adjacent}{hypotenuse}cosα=

hypotenuse

adjacent

, and solve for the distance between the Sun and the shooting star "AC":

cos(x\°)=\frac{y}{AC}cos(x\°)=

AC

y

AC=\frac{y}{cos(x\°)}AC=

cos(x\°)

y

7 0
3 years ago
Based on the graph, what is the initial value of the linear relationship?
yuradex [85]

Answer:

The initial value of the linear relationship is y=-2

Step-by-step explanation:

we know that

The y-intercept is the value of y when the value of x is equal to zero

The initial value is the value of the linear function when the value of x is equal to zero

therefore

The initial value is the y-coordinate of the y-intercept of the linear function

Observing the graph

The y-intercept is the point (0,-2)

so

For x=0

The value of y is equal to y=-2



6 0
3 years ago
Read 2 more answers
Select all of the factors of x3 + 5x2 + 2x – 8.
Nikolay [14]

Answer:

i think it would be multiplied by x

Step-by-step explanation:

that sounds like the best thing to do

5 0
3 years ago
Read 2 more answers
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