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3 years ago

5

The given triangles are similar. Solve for x.

1 answer:

Maru [420]3 years ago

8 0

The first one is 21.93 rounded up to 22.

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A triangular pyramid and its net are shown. Rain uses Calculations below to conclude that the surface area of a triangular pyram

Harman [31]

Answer:

Rain is correct, calculates correctly, 238.5 square inches

Step-by-step explanation:

The given triangular pyramid is shown with the dimensions.

The surface area of the triangular pyramid is given by :

$A=\frac{1}{2}\left(a \times b\right) + 3\left(\frac{1}{2}\times b\times s\right)$

$A=\frac{1}{2}\left(10 \times 8.7\right) + 3\left(\frac{1}{2}\times 10\times 13\right)$

$A=43.5+(3\times 65)$

$A=43.5+195$

$A=238.5 \ \text{inche}^2$

Therefore, the surface area of the triangular pyramid is = 238.5 square inches.

5 0

2 years ago

12• (-9) pls help and explain to mr

Paha777 [63]

12*-9 is simply 12 * 9 with a negative at the end. I use a strategy as I'm in 8th grade that I learned in 5th this should help. If you love to hat that is bad. this is for division and multiplication this is stating that if a positive( love ) number times or divided by a negative( hate ) number it is negative( bad ). so 12*9=108 and add the negative and your answer is -108
Hope this helps:) good luck!

3 0

2 years ago

Danielle plans on making monthly payments into an account that earns 2.5% compounded monthly. If she wants $10,000 in 22 years,

Nesterboy [21]

Answer:

5,500

Step-by-step explanation:

I=PRT/100

Then

I = 2.5*10,000*22/100

I=#5,500

4 0

2 years ago

Can someone please help me?

nataly862011 [7]

Answer:

Step-by-step explanation:

7⁴/ 7⁶ = 1/ 7⁶⁻⁴ = 1/7² = 1 /49 = 0.02

6 0

3 years ago

Read 2 more answers

Using mathematical induction prove whether or not the following statement is true for all positive integers n, or show why it is

aniked [119]

Base case: For $n=1$ , the left side is 2 and the right is $2\cdot1^2=2$ , so the base case holds.

Induction hypothesis: Assume the statement is true for $n=k$ , that is

$2+6+10+\cdots+4k-2=2k^2$

We want to show that this implies truth for $n=k+1$ , that

$2+6+10+\cdots+4k-2+4(k+1)-2=2(k+1)^2$

The first $k$ terms on the left reduce according to the assumption above, and we can simplify the $k+1$ -th term a bit:

$\underbrace{2+6+10+\cdots+4k-2}_{2k^2}+4k+2$

$2k^2+4k+2=2(k^2+2k+1)=2(k+1)^2$

so the statement is true for all $n\in\mathbb N$ .

5 0

3 years ago