MATHEMATICS HIGH SCHOOL

The perimeter of a rectangular garden is 338 m.If the width of the garden is 74 m, what is its length?

Answers

Answer 1
Answer:

Answer:

l=95

Step-by-step explanation:

Perimeter is all the sides add up. So the equation if you know two of the sides, it should be: 338=(2*74)+(2*l)
Answer 2
Answer:

Answer:

Step-by-step explanation:

the perimeter is 2w plus 2l

so 338 = 74(2)+2l

338 = 148+2l

190=2l

l=95

your length would be 95 m

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What is the slope of a line parallel to y = -5x + 2?

Answers

Answer:

-5

Step-by-step explanation:

Parallel lines have the same slope, so it would just be -5

Will give brainliest!!Can someone please explain how i do problems like these my teacher makes no sense

Select the correct answer.
Which expression is equivalent to the given expression?
(12g^2h^7)(3g–^2h-4)

A. 36gh^3

B. 36h^3

C. 36/g^4h^28

D. 38/h^3

Answers

Answer:

C

Step-by-step explanation:

Answer:

its C

Step-by-step explanation:

Which of the following shows the polynomial below written in descending order?4x2 - x + 8x6 + 3 + 2x10

A. 2x10 + 8x6 + 4x2 - x + 3
B. 8x6 + 4x2 + 3 + 2x10 - x
C. 3 + 2x10 + 8x6 + 4x2 - x
D. 3 - x + 2x10 + 8x6 + 4x2

Answers

The correct awnser should be A cause for them to be decending the biggest variable goes first and then works its way down to the number without and variable.


A is the correct awnser
A indeed is the correct answer

Each of the 20 members of a club needs to purchase a t-shirt and a flag. T-shirts cost $10 each. Flags cost $2.50 each. Before tax, what is the cost for all the t-shirts and flags?

Answers

Answer: $250.00

Step-by-step explanation:

1 Member needs one t-shirt and on flag.

$10.00+$2.50=$12.50 per member.

$12.50x20 members=$250.00

In a sample of 50 households, the mean number of hours spent on social networking sites during the month of January was 45 hours. In a much larger study, the standard deviation was determined to be 8 hours. Assume the population standard deviation is the same. Which of the statement below best describes that there is a 95% confidence interval for the mean hours devoted to social networking in January? A. The 95% confidence interval ranges from 8 to 45 hours. B. The 95% confidence interval ranges from 40.13 to 45.78 hours. C. The 95% confidence interval ranges from 43.87 to 46.13 hours. D. The 95% confidence interval ranges from 42.78 to 47.22 hours.

Answers

Answer:

D. The 95% confidence interval ranges from 42.78 to 47.22 hours.

Step-by-step explanation:

In a sample of 50 households, the mean number of hours spent on social networking sites during the month of January was 45 hours. In a much larger study, the standard deviation was determined to be 8 hours.

Here,

n = sample size = 50,

μ = mean = 45,

σ = standard deviation = 8,

We know that, confidence interval will be,

=\mu\ \pm\ z(\sigma )/(√(n))

For a confidence interval of 95%, we use z = 1.96, putting the values

=45\ \pm\ 1.96(8)/(√(50))

=42.78,47.22
D. The 95% confidence interval ranges from 42.78 to 47.22 hours.

To solve the interval, we used the upper and lower limit formulas wherein:
xbar = 45
z = 1.96
s = 8
n = 50

Solve the equation by taking the square root 4(x+1)^2=100

Answers

Answer:

Step-by-step explanation:

4(x+1)²=100

(x+1)²= 5²

(x+1)²- 5² = 0

(x+1-5)(x+1+5) = 0  by identity : a² - b ² = (a - b) (a+b)

(x - 4 ) (x+6) =0

x - 4 = 0 or x+6 = 0

x = 4  or x = - 6
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