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Lana71 [14]
3 years ago
5

What is a part of a line , has one endpoint ,continues in one direccion. Math homework

Mathematics
1 answer:
kicyunya [14]3 years ago
5 0
A ray is a line has one endpoint and a line that goes on forever.
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Find the equation of the line that is perpendicular to y = 2x + 8 and passes though the point (8, –8).
irina1246 [14]

Answer:

y= -1\2x-4

Step-by-step explanation:

y=mx+c

perpinicular means you flip the slop and change the sign so the slope(m) become -1\2

y=-1\2x+c

since he gives you (8,-8) meaning x=8 and y= -8

we substitute

-8=(-1\2 x 8) +c

now we isolate c

-8+4=c

c=-4

the equation will be

y= -1\2x-4

3 0
3 years ago
Which are true about the area of a circle? Check all that apply
Zigmanuir [339]

Answer:

The units are always squared. The area can be found when given the diameter by first finding the radius. Circumference can be used to find area.

Step-by-step explanation:

7 0
2 years ago
A chocolate chip cookie recipe calls for 6 cups of chopped nuts and 8 cups of chocolate chips. When Gregory made a batch of thes
Radda [10]

Answer:

He did not

Step-by-step explanation:

if you put 6: 8 in a fraction 6/8, it doesn't have any relationship with 9/12

3 0
3 years ago
See attachment below
KatRina [158]

Answer: Third Option

x=1.469743

Step-by-step explanation:

We have the following exponential equation

3^{x+1}=15

We must solve the equation for the variable x

To clear the variable x apply the log_3 function on both sides of the equation

log_3(3^{x+1})=log_3(15)

Simplifying we get the following:

x+1=log_3(15)

To simplify the expression log_3 (15) we apply the base change property

log_b(y)=\frac{log(y)}{log(b)}

This means that:

log_3 (15)=\frac{log(15)}{log(3)}

Then:

x+1=\frac{log(15)}{log(3)}

x=\frac{log(15)}{log(3)}-1

x=1.469743

6 0
2 years ago
On a unit circle, the vertical distance from the x-axis to a point on the perimeter of the circle is twice the horizontal distan
swat32

Check the picture below.

since the vertical distance, namely the y-coordinate, is twice as much as the horizontal, then if the horizontal is "x", the vertical one must be 2x.

let's find the hypotenuse first.

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}

5 0
3 years ago
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