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3 years ago

HELP PLEASE find the surface area of the prism

Mathematics

1 answer:

attashe74 [19]3 years ago

7 0

Answer:

9:130 10:198 11:76

Step-by-step explanation:

S.A.:= 2LW+2LH+2WH

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If your cell phone bill was 67 Point $82 and there is a 7.5% late fee how much will your bill be with the late fee included

Ivenika [448]

$\frac{7.5}{100} = \frac{x}{82} \\ x = \frac{82 \times 7.5}{100} = \frac{615}{100 } = 6.15$
$82 + 6.15 = 88.15$
the answer is $88.15

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3 years ago

there are eight competitors in each ring for a raw kwon do to tournament if there are 96 competitors in the tournament how many

kvv77 [185]

Answer:

Step-by-step explanation:

You divide 96 by 8 and you get 12.

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2 years ago

Find the quotient. Write your answer as a decimal. 12.42 divided by (-4.8) =

MakcuM [25]

Answer:

Multiply the divisor by a power of 10 to make it a whole number.

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Divide the dividend by the whole-number divisor to find the quotient.

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2 years ago

What is the equation of the line shown in the coordinate plane below?

valentina_108 [34]

Answer:

There's no picture for us to refer to. I need a picture to answer the question.

3 0

2 years ago

Read 2 more answers

A. f(x) = l-2xl - 3 Domain: {1, 2, 3} Find the range

gizmo_the_mogwai [7]

Hello there!

We are given the function:

$\displaystyle \large{ f(x) = | - 2x| - 3} \\$

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

x = 1

$\displaystyle \large{ f(1) = | - 2(1)| - 3} \\ \displaystyle \large{ f(1) = | - 2| - 3} \\$

Recall that any numbers in absolute value are always positive.

$\displaystyle \large{ f(1) = 2 - 3} \\ \displaystyle \large{ f(1) = - 1} \\$

x = 2

$\displaystyle \large{ f(2) = | - 2(2)| - 3} \\ \displaystyle \large{ f(2) = | - 4 | - 3} \\ \displaystyle \large{ f(2) = 4 - 3} \\ \displaystyle \large{ f(2) = 1}$

x = 3

$\displaystyle \large{ f(3) = | - 2(3)| - 3} \\ \displaystyle \large{ f(3) = | - 6| - 3} \\ \displaystyle \large{ f(3) = 6 - 3} \\ \displaystyle \large{ f(3) = 3}$

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

6 0

3 years ago