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VashaNatasha [74]
3 years ago
12

PLSS HELP ILL GIVE BRAINLIEST IF YOU HELP ME AND EXPLAIN UR ANSWER

Mathematics
2 answers:
DanielleElmas [232]3 years ago
7 0

Answer:

h = V/πr²

Step-by-step explanation:

Its asking you to isolate h. You can simply divide on the right besides h on both sides.

V = πr²h

V/πr² = h

kherson [118]3 years ago
3 0

Answer:

It is the second choice.

Step-by-step explanation:

Divide both side by pi r squared. Boom.

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Answer:

The first statement: "Her first mistake was in Step 2. She added 0.2 to each term instead of multiplying by 0.2."

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3 years ago
The two polygons are similar. Find the values of x and y. <br> Please help, i will mark brainliest
9966 [12]
Do you have a photo?
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3 years ago
Find the x and y intercept of y=7×+5
Zina [86]
X-int. is (5/7,0)
y-int. is (0,5)
3 0
4 years ago
How does f(x) = 3* change over the interval x = 4 to x = 57
Vilka [71]

f(x) increase by a factor of 3

Explanation:

        Given that f(x)= 3* and the interval is x=4 to x=57

        Now we put the value for x is 4 to 57 then value of f(x) increase with the multiply of 3.

        Because the x is multiplied with 3 i.e., 3*

        So f(x) increase by a factor of 3.

        If we put x=4, then f(x)= 12     (∵ 3×4=12)

        If we put x=5, the f(x)= 15       (∵ 3×5=15)

        If we put x=6,the f(x)= 18             (∵ 3×6=18)

similarly., values of x= 7,8,9,...155.

        Then,

        If we put x=56, the f(x)=168

        This process will continue until f(x)=171 for x=57.

6 0
3 years ago
Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the
Ivan

Let L be the line given by the vector equation

(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}.

First, we use the director vectors of the lines L1 and L2 to get the

vector equation of the plane containing them, which we denote by \Pi_1. This is,

\\\\\Pi_1  : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\

We observe that \vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0). Therefore, the vector equation of \Pi_1 defines a plane and \vec{N} is a normal vector to \Pi_1.

 

Finally, the vector equation for the wanted plane, which we denote by \Pi, is

\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .

Thus, if s = 0, then L \subset \Pi and since \vec{N} is parallel to \Pi, then it is perpendicular to \Pi_1.

8 0
3 years ago
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