All categories

3 years ago

What is the area of this face? A.150in2 B.100in2 C.400in2 D.200in2

Mathematics

1 answer:

77julia77 [94]3 years ago

8 0

Answer:

it's

$200 {in}^{2}$

Step-by-step explanation:

$10 \times 10 = \: \: \: \: \: 100 \\ 10 \times 10 = + 100 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= 200 {in}^{2}$

You might be interested in

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

4 years ago

SELECT ALL: The point (0, 10) is a solution to the following system of inequalities. Given that information, what could be the s

r-ruslan [8.4K]

63is the number Sorry I don’t know

8 0

3 years ago

Needddd helpppppp !!!

ser-zykov [4K]

B. I’m pretty sure. Since the arrow is pointing towards the S which is at the beginning.

3 0

3 years ago

Read 2 more answers

A college football coach has decided to recruit only the heaviest 15% of high school football players. He knows that high school

Alla [95]

Answer:

The coach should start recruiting players with weight 269.55 pounds or more.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 225 pounds

Standard Deviation, σ = 43 pounds

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.15

$P( X > x) = P( z > \displaystyle\frac{x - 225}{43})=0.15$

$= 1 -P( z \leq \displaystyle\frac{x - 225}{43})=0.15$

$=P( z \leq \displaystyle\frac{x - 225}{43})=0.85$

Calculation the value from standard normal z table, we have,

$P(z < 1.036) = 0.85$

$\displaystyle\frac{x - 225}{43} = 1.036\\\\x = 269.548 \approx 269.55$

Thus, the coach should start recruiting players with weight 269.55 pounds or more.

3 0

4 years ago

Someone help me please, 25 points

ycow [4]

A. (x+8)^ 2 + 1
B. (x-3)^ 2 - 8
C. ( x - 3/2) ^ 2 - 1/4

4 0

2 years ago