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MrRissso [65]
2 years ago
12

ILL GIVE BRAINLIEST ON THIS QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
DochEvi [55]2 years ago
8 0

Answer:

15.25 or 15 if you need it to be rounded.

Step-by-step explanation:

-11 on both sides to get 5x=1x+61. Then, -1x on both sides to get 4x=61. Then to find x you do 61 divided by 4 which is 15.25.

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You will get points for asking me questions Ask me any question and ill answer it
vampirchik [111]

A large soft drink costs one and a half times as much as small soft drink. If a small drink costs n dollars, write an expression representing how much a large drink costs.

A) 1.5n

B) 1.5 + n

C) 0.5 + n

D)

1

__   n

2




3 0
3 years ago
2. The point (0.28, 0.96) lies on the unit circle. Which of the following is closest to the tangent of an angle
vlada-n [284]

Answer:

(1) 3.43

Step-by-step explanation:

tan θ = y / x

tan θ = 0.96 / 0.28

tan θ = 3.43

7 0
3 years ago
Solve using elimination.<br> 5x - 6y = 2<br> -5x - 5y = 20
s344n2d4d5 [400]

Answer:

x = -2, y =-2

Step-by-step explanation:

5x - 6y = 2

-5x - 5y = 20

Add the two equations together to eliminate x

5x - 6y = 2

-5x - 5y = 20

------------------------

-11y = 22

Divide by -11

-11y/-11 = 22/-11

y = -2

Now we can find x

5x -6(-2) =2

5x +12 = 2

Subtract 12 from each side

5x +12-12 =2-12

5x = -10

Divide by 5

5x/5 = -10/5

x = -2

6 0
3 years ago
the Vertices of a hyperbola are at (-5, -2) and (-5, 12) and the point (-5, 30) is one of its foci. What is the equation of its
SpyIntel [72]
With the provided vertices and the focus point, the hyperbola will look like the one in the picture below.

notice the "c" distance from the center to the focus point.

since it's a vertical hyperbola, the positive fraction will be the one with the "y" in it, its center is clearly half-way between the vertices at -5, 5.

its major axis or traverse axis goes from -2 up to 12, so is 14 units long, therefore the "a" component is half that, or 7.

\bf \textit{hyperbolas, vertical traverse axis }&#10;\\\\&#10;\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ( h, k)\\&#10;vertices\ ( h,  k\pm a)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{ a ^2 + b ^2}&#10;\end{cases}\\\\&#10;-------------------------------

\bf \begin{cases}&#10;h=-5\\&#10;k=5\\&#10;a=7\\&#10;c=25&#10;\end{cases}\implies \cfrac{(y-5)^2}{7^2}-\cfrac{[x-(-5)]^2}{b}=1&#10;\\\\\\&#10;\cfrac{(y-5)^2}{7^2}-\cfrac{(x+5)^2}{b}=1\\\\&#10;-------------------------------\\\\&#10;c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\implies \sqrt{25^2-7^2}=b&#10;\\\\\\&#10;\sqrt{625-49}=b\implies \sqrt{576}=b\implies 24=b\\\\&#10;-------------------------------\\\\&#10;\cfrac{(y-5)^2}{7^2}-\cfrac{(x+5)^2}{24^2}=1\implies \cfrac{(y-5)^2}{49}-\cfrac{(x+5)^2}{576}=1

8 0
3 years ago
Can somebody help me?
nata0808 [166]
Answer: I believe the answer would be -2 so plot it on -2.

Explanation: Point P is -2.5 and point R is 0.5. Adding those would equal -2.

i hope this helps! :)
7 0
3 years ago
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