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Nataly [62]
3 years ago
13

Piper claims that 3(2a + 2) is equivalent to 5a + 6. She checks her claim by substituting -3 for a. Is Piper's claim correct? Wh

y or Why not?
Please Help SOS!!!
Mathematics
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer:

Piper's claim that 3(2a+2) is equal to 5a +6 is incorrect.

Step-by-step explanation:

The first step is to use the distributive property to distribute the 3 to BOTH the 2a and 2. The resulting expression is 6a+6 NOT 5a+6. Piper's mistake was most probably that she added 3 and 2 to get 5 instead of multiplying them together to get 6.

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Classify this triangle by its sides?
mel-nik [20]

Answer:

Isosceles

Step-by-step explanation:

Hope this helps you

4 0
3 years ago
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Solve for x. 9 4 13.5 x+4
Lady_Fox [76]

Answer:

x = 2

Step-by-step explanation:

The parallel lines divide the transversals into proportional segments, that is

\frac{x+4}{13.5} = \frac{4}{9} ( cross- multiply )

9(x + 4) = 54 ( divide both sides by 9 )

x + 4 = 6 ( subtract 4 from both sides )

x = 2

5 0
3 years ago
An event lasts 4 hours and 20 minutes. How long did the event last in seconds?
enyata [817]
20*6=120seconds +3600second in an hour so 3600*4=14400

1210+14400=14520

14520 is the answer I hope this helps!


8 0
3 years ago
Read 2 more answers
Can someone and give me the right answer
kolezko [41]

The 1st option is your answer,

because to solve c, the -11 and 11 = 0, so your left with adding 23 and 11, then you divide -4 on both sides to get the variable (x) by itself

And for f, you only need to divide -4 on both sides to get x by itself.

6 0
3 years ago
BRAINLIESTT ASAP! PLEASE HELP ME :)
Maslowich

Step-by-step explanation:

\displaystyle y = Asin(Bx - C) + D

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:

\displaystyle Phase\:[Horisontal]\:Shift → \frac{0}{\frac{1}{7}} = 0 \\ Period → \frac{2}{1}π = 2π

Therefore we have our answer.

Extended Information on the trigonometric function

\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|

NOTE: Sometimes, your <em>vertical shift</em> might tell you to shift your graph below or above the <em>midline</em> where the amplitude is.

I am joyous to assist you anytime.

3 0
3 years ago
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