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2 years ago

Find an expression which represents the sum of (3x+2)(3x+2) and (6x-10)(6x−10) in simplest terms.

Mathematics

2 answers:

ArbitrLikvidat [17]2 years ago

4 0

Step-by-step explanation:

sum means add

(3x+2)(3x+2) = 6x+4

kenny6666 [7]2 years ago

4 0

Here is the answer. Hope this helps!

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There is 174 games scheduled for 6 different leagues each league has the same number of games scheduled how many games does each

maxonik [38]

Twenty nine games in each league

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3 years ago

A group of 18 people ordered soup and sandwiches for lunch. Each person in the group had either one soup or one sandwich. The sa

IceJOKER [234]

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3 years ago

10g - 4 (g - 2) + 11 = 1

sergij07 [2.7K]

Answer:

g = -3

Step-by-step explanation:

10g-4(g-2)+11=1

1) 10g-4g+8+11=1 Combine like terms

2) 6g+8+11=1

3) 6g+19=1

4) 6g=-18 To continue the process of getting the variable by itself, we have to get rid of 19. So we are going to subtract 19 from 19 to get 0, and whatever we do to the other side, so we subtract 19 from 1 to get -18.

5)6g=-18 To get the variable by itself, instead of mutiplying, we will divide by 6 on both sides to get... g=-3! Hope this helps!

5 0

3 years ago

A recipe includes 8 cups of flour and 2/3 third cups of sugar right the ratio of the amount of flour to the amount of sugar as a

Art [367]

Its 8:2/3, we can write it as $\frac{8}{ \frac{2}{3} } =8 *\frac{3}{2} = \frac{8*3}{2} = \frac{24}{2}= \frac{12}{1}$
Now we can back to 12:1 - its the result

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3 years ago

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard

bulgar [2K]

We have been given that the lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. We are asked to find the probability of a lion living longer than 10.1 years using empirical rule.

First of all, we will find the z-score corresponding to sample score 10.1.

$z=\frac{x-\mu}{\sigma}$ , where,

z = z-score,

x = Random sample score,

$\mu$ = Mean

$\sigma$ = Standard deviation.

$z=\frac{10.1-12.5}{2.4}$

$z=\frac{-2.4}{2.4}$

$z=-1$

Since z-score of 10.1 is $-1$ . Now we need to find area under curve that is below one standard deviation from mean.

We know that approximately 68% of data points lie between one standard deviation from mean.

We also know that 50% of data points are above mean and 50% of data points are below mean.

To find the probability of a data point with z-score $-1$ , we will subtract half of 68% from 50%.

$\frac{68\%}{2}=34\%$

$50\%-34\%=16\%$

Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.

3 0

3 years ago