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Over [174]
1 year ago
15

ANSWERS ASAP PLEASEEE

Mathematics
1 answer:
Lorico [155]1 year ago
7 0
Y = 4x - 9

I hope this helps
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one vertex of a triangle is located at (0,5) on a coordinate grid after a transformation the vertex is located at(5,0) which tra
Alik [6]

Answer:

It's a rotation clockwise of 90 degrees about the origin.

Step-by-step explanation:

The given point moves from   y = 5 to x = 5 so it has passed through 90 degrees about the origin.

It's a rotation clockwise of 90 degrees about the origin.

6 0
3 years ago
Find the area of a circle with a circumference of 31.5 units
Westkost [7]

Answer:

79 units²

Step-by-step explanation:

The area of a circle is found by using the formula A = π x r². Since we are not given the radius we can use the formula r = C / (2π). Substituting in the given values, we get r = 31.5/(2π). 2 times pi is 6.283185307. 31.5 divided by this will get you 5.013380707. Now, put that value in for r in the area equation. A = π x 5.013380707². Pi times 5.013380707² equals 78.96074613. This can be rounded to 79. (Or 80 if you need to the nearest whole number.)

6 0
3 years ago
Idk the answer to this question please help me
erica [24]
2/3 x 42/3
hope this helps

7 0
4 years ago
I need help with 3 problems of Properties of Signed Numbers somebody please help me
Shalnov [3]
1. 2/3 - (1/4)*(4/9)
     2/3-4/36
     24/36-4/36
     20/36 also same as 4/9 (D)


2. It's B. (-3*(2*7)=(-3*2)*7

3. -4+(-6)*(-3)
     -4+18
      -14 (b)


7 0
3 years ago
a circular oil slick is expanding at a rate of 2m^2/h. Find the rate at which it's diameter is expanding when it's radius is 1.5
aleksandrvk [35]

Answer:  \frac{4\pi}{3} \text{ meter per hour}

Step-by-step explanation:

The circular oil slick is expanding at a rate of  2 m^2/h

Let A be the area of the circular oil slick,

So, the changes in  A with respect to time (t),

\frac{dA}{dt} = 2

\frac{d(\pi r^2)}{dt} = 2

2\pi r\frac{dr}{dt} = 2

\frac{dr}{dt} = \frac{1}{\pi r}  

Also, the change in diameter with respect to time(t),

\frac{d}{dt} (2 r) = 2 \frac{dr}{dt}

\frac{d}{dt} (2 r) = 2 \times \frac{1}{\pi r}

\frac{d}{dt} (2 r) = \frac{2}{\pi r}

For r = 1.5 m,

\frac{d}{dt} ( 2 r)]_{r=1.5} = \frac{2}{\pi \times 1.5}=\frac{20}{\pi \times 15}=\frac{4\pi }{3}\text{ meter per hour}




7 0
3 years ago
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