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3 years ago

How many solutions does 6x+ m15=6(x-3)

Mathematics

2 answers:

alexgriva [62]3 years ago

8 0

Answer:

Step-by-step explanation:

Hey evrbody how y’all doing

Andre45 [30]3 years ago

6 0

Answer:

Step-by-step explanation:

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A sector of a circle of 7cm has an area of 44cm².calculate the angle of the sector,correct to the nearest degree π =22\7

Hitman42 [59]

Answer:

$\theta=102.85^{\circ}$

Step-by-step explanation:

Given that,

The radius of circle, r = 7 cm

The area of sector, A = 44cm²

We need to find the angle of the sector. The formula for the area of sector is given by :

$A=\dfrac{\theta}{360}\pi r^2$

Solve for $\theta$ .

$\theta=\dfrac{360A}{\pi r^2}\\\\\theta=\dfrac{360\times 44}{\dfrac{22}{7}\times 7^2}\\\\\theta=102.85^{\circ}$

So, the angle of the sector is equal to $102.85^{\circ}$ .

7 0

2 years ago

Find the range of the function. ƒ(x) = x2 + 3

beks73 [17]

Answer:

Step-by-step explanation:

7 0

2 years ago

K=mv^2/2 solve for v

matrenka [14]

Answer: $v=\sqrt[]{\frac{2K}{m} }$

Step-by-step explanation:

$K=\frac{mv^2}{2}$

First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;

$2*K=\frac{mv^2}{2}*2$

$2K=mv^2$

Divide by m to isolate $v^2$ .

$\frac{2K}{m}=\frac{mv^2}{m}$

$\frac{2K}{m} =v^2$

To eliminate the square and isolate v, extract the square root.

$\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}$

$\sqrt[]{\frac{2K}{m} }=v$

let's rewrite it in a way that v is in the left side.

$v=\sqrt[]{\frac{2K}{m} }$

4 0

2 years ago

Help i cant figure this out:(

stealth61 [152]

Answer:

6-2=4

31/5 x 8 = 248/40

15/8 x 5 = 75/40

248-75=173

4 13/40

answers in order

6-2=4

6 1/5 - 1 7/8 = 4 13/40

Hope this helps

Step-by-step explanation:

8 0

2 years ago

Simplify and expand the equation below <br>7(2-x)-(x-3)

Tanya [424]

Answer:

- 8x + 17

Step-by-step explanation:

7(2 - x) - (x - 3)

14 - 7x - x + 3

- 8x + 14 + 3

- 8x + 17

3 0

2 years ago