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2 years ago

Complete the steps to find the value of x.

Mathematics

2 answers:

k0ka [10]2 years ago

8 0

Answer:

Where is the equation for me to solve for x?

Step-by-step explanation:

olga55 [171]2 years ago

4 0

Answer: you have to GIVE ME THE EQUATION/PROBLEM TO TELL YOU

Step-by-step explanation: So bad man

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What is the circumference of a circle with a radius of 5 m?

Lena [83]

Answer:

31.4 meters

Step-by-step explanation: use the formula 2 times pi times 5 then you just plug in 5 into the radius

5 0

3 years ago

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I need help with 17-22

DIA [1.3K]

Answer:

-5

Step-by-step explanation:

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3 years ago

Evaluate the expression if m = 3.<br><br> 3m^2<br><br><br> (SHOW YOUR WORK)

iragen [17]

The value of given expression when m = 3 is 27

<h3><u>Solution:</u></h3>

Given expression is $3m^2$

We have to evaluate the given expression for m = 3

To find for m is equal to 3, substitute m = 3 in given expression

From given expression,

$\rightarrow 3m^2$

Plug in m = 3 in above expression

$\rightarrow 3(3)^2$ ------ eqn 1

We know that,

$a^2$ can be expanded as,

$a^2=a \times a$

Applying this in eqn 1, we get

$\rightarrow 3(3)^2=3 \times (3 \times 3)$

Simplify the above expression

$\rightarrow 3(3)^2=3 \times (3 \times 3) = 3 \times 9 = 27$

Therefore, for m = 3 we get,

$\rightarrow 3(3)^2=27$

Thus value of given expression when m = 3 is found

5 0

3 years ago

Solve the following by using mathematical Induction. For >/ 1

cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

$1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

1. When $n=1,$ we have

in left part $1^2=1;$

in right part $\dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.$

2. Assume that for all $k$ following equality is true

$1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)$

3. Prove that for $k+1$ the following equality is true too.

$1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)$

Consider left part:

$1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}$

Consider right part:

$\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)$

We get the same left and right parts, so the equality is true for $k+1.$

By mathematical induction, this equality is true for all n.

3 0

3 years ago

Help!!!!! please!!!!!

GaryK [48]

Answer: D) 602.88

Step-by-step explanation:

A cylinder is basically a circle prism. Any prism volume can be found by taking the area of the base * the height. The circle area is 4^2*3.14 = 50.24. Then multiply that by the height(12) 50.24*12 = 602.88

Also, so far all of your questions that I've answered have been D...

Hope it helps <3

6 0

2 years ago

Read 2 more answers