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FrozenT [24]
2 years ago
6

12/02/2020 someone help me

Mathematics
1 answer:
krek1111 [17]2 years ago
6 0

Answer:

C. h = \frac{V}{\pi r^{2} }

Step-by-step explanation:

Since V = \pi × r × h, h = V ÷ (\pi × r) because the × sign changes to a ÷ sign when h is removed and V is moved from the left side of the equation to the right side.

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cricket20 [7]

\qquad\qquad\huge\underline{{\sf Answer}}

Here's the solution :

9. Statement : \tt{VB\:\: bisects\;\; \angle EVO}

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  • Reason : \tt Definition \;\; of \;\; angle \;\; bisector

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  • Reason : \tt Given

12. Statement :\tt \angle1 \cong \angle 2

  • Reason : \tt Definition \;\; of \;\; angle \;\; bisector

13. Statement : \tt \overline{BV }\cong \overline{BV}

  • Reason : \tt Common\:\: side

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7 0
2 years ago
Cot0 + tan0 = sec0csc0
Lelechka [254]

Answer:

??

Step-by-step explanation:

7 0
3 years ago
The graph below represents the solution set of which inequality?
natulia [17]

Answer:

option: B (x^2+2x-8) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

x^2-2x-8

On using the method of splitting the middle term we have:

x^2-4x+2x-8

⇒  x(x-4)+2(x-4)=0

⇒ (x+2)(x-4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x+2>0 and x-4

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

x+2 and x-4>0

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

x^2+2x-8

On using the method of splitting the middle term we have:

x^2+4x-2x-8

⇒ x(x+4)-2(x+4)

⇒ (x-2)(x+4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x-2>0 and x+4

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

x-2 and x+4>0

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

x^2-2x-8>0

On using the method of splitting the middle term we have:

x^2-4x+2x-8>0

⇒ x(x-4)+2(x-4)>0

⇒ (x-4)(x+2)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x+2>0 and x-4>0

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

x+2 and x-4

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

x^2+2x-8>0

On using the method of splitting the middle term we have:

x^2+4x-2x-8>0

⇒ x(x+4)-2(x+4)>0

⇒ (x-2)(X+4)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x-2 and x+4

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

x-2>0 and x+4>0

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




5 0
3 years ago
Complete the ordered pair for the given equation. (_,-1);y=-3/5x-7
Fiesta28 [93]
Hi!

So for this problem, we're given the y-coordinate and we can plug that in for y in the equation. So let's do that

-1 = -3/5x - 7
6 = -3/5x
30 = -3x
-10 = x

The missing coordinate would be x, assuming that the equation was y = (-3/5)x - 7

3 0
3 years ago
Hello please help i’ll give brainliest
Ivenika [448]

Answer:

Step-by-step explanation:

Dot plot

8 0
2 years ago
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