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2 years ago

Simplify by combining like terms: 3 1/9p + 5 - 1/3p

Mathematics

2 answers:

diamong [38]2 years ago

8 0

Answer:

$7\frac{2}{3}$

Step-by-step explanation:

$3\frac{1}{9}p+5-\frac{1}{3}p$

Put it into a different form to make it easier.

$3\frac{1}{9}p-\frac{1}{3}p+5$

Add the fractions. We need to figure out the LCM to figure out the common denominator. The LCM is 9 in this case.

$3\frac{1}{9}-\frac{3}{9} +5$

I like to convert into imroper fractions. This is optional though.

$\frac{28}{9} -\frac{3}{9} +5$

$\frac{24}{9} +5$

$\frac{8}{3}+5$

Get a common denominator. The LCM is 3.

$\frac{8}{3}+\frac{15}{3}$

$\frac{23}{3}$

Make this fraction into a mixed number.

$7\frac{2}{3}$

I hope this helps!

Airida [17]2 years ago

6 0

Answer:

5(5p+9)

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Step-by-step explanation:

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3 years ago

Solve for c. −58c=20 A-34 B-32 C26 D28

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The answer is A: -34.

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3 years ago

-4 plus the square root of 81

almond37 [142]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ The square root of 81 is 9

We have -4 + 9

Thus, your answer is 5

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3 years ago

Read 2 more answers

Find three consecutive integers whose sum is equal to −354. Three consecutive integers in ascending form are __,__,__

mixer [17]

Answer: -119, -120, and -121

Step-by-step explanation: This problem states that the sum of three consecutive integers is -354 and it asks us to find the integers. Three consecutive integers can be represented as followed.

X ⇒ first integer

X + 1 ⇒ second integer

X + 2 ⇒ third integer

Since the sum of our three consecutive integers is -354, we can set up an equation to represent this.

X + X + 1 + X + 2 = -354

Now, we can simplify on the left side of the equation.

3x + 3 = -354

-3 -3 ← subtract 3 on both sides of the equation

3x = -357

÷3 ÷3 ← divide both sides by 3

X = -119

X ⇒ first integer = -119

X + 1 ⇒ second integer = -120

X + 2 ⇒ third integer = -121

Therefore, our three consecutive integers are -119, -120, and -121.

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3 years ago

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Sam entered three functions into a graphing calculator, y1 = x + 2, y2 = x2 + 2, and y3 = 2x. A portion of the table created by

V125BC [204]

For this case we have the following functions:
$y1 = x + 2

y2 = x ^ 2 + 2

y3 = 2 ^ x$

When x = 0 we have:
For y1:
$y1 = 0 + 2

y1 = 2$
For y2:
$y2 = 0 ^ 2 + 2

y2 = 0 + 2

y2 = 2$
Therefore, we have to:
$y1 = y2
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When x = 5 we have:
For y2:
$y2 = 5 ^ 2 + 2

y2 = 25 + 2

y2 = 27$
For y3:
$y3 = 2 ^ 5

y3 = 32$
Therefore, we have to:
$y2 \ \textless \ y3
$

When x = -1 we have:
For y1:
$y1 = -1 + 2

y1 = 1$
For y2:
$y2 = (-1) ^ 2 + 2

y2 = 1 + 2

y2 = 3
$
For y3:
$y3 = 2 ^ {-1}

y3 = 1/2

y3 = 0.5$
Therefore, we have to:
$y3 \ \textless \ y1 \ \textless \ y2
$

Answer:
When x = 0, y1 = y2
When x = 5, y2 <y3
When x = -1, y3 <y1 <y2

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3 years ago