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Maurinko [17]
3 years ago
15

Please help with question with explanations. Ty.

Mathematics
1 answer:
earnstyle [38]3 years ago
4 0

Answer:

last option is correct answer

Step-by-step explanation:

perimeter = 2(l + w) \\  = 2( \frac{5x + 10}{2x + 5}  +  \frac{x - 2}{2x + 5} ) \\  = 2( \frac{5x + 10 + x - 2}{2x + 5} ) \\  = 2( \frac{6x + 8}{2x + 5} ) \\  =  \frac{12x + 16}{2x + 16}

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The spinner below is spun once and a card is drawn from a deck of 4 cards labeled A, B, C, and D. Find the probability of each e
Lorico [155]

Answer:

the prob of a,b,c or d being picked up is 1/4

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the value of x. then tell whether the side lengths form a Pythagorean triple
timurjin [86]

Answer:

x = √226

Step-by-step explanation:

Pythagorean Theorem: a² + b² = c²

In that case, plug in <em>a </em>and <em>b</em>,

15² + 1² = c²

c = √(15² + 1²)

c = 15.0333

This is NOT a Pythagorean triple.

5 0
3 years ago
what is an art and a science that deals with the collection,organization,creative presentation and analysis of data? A.physics B
DENIUS [597]

Answer:

The answer is B. Statistics

7 0
3 years ago
Can you solve these problem and show work for each one of these please.
pychu [463]
Consider this option.
Change the design according to the local requirements.

8 0
3 years ago
Does anyone know how to do this help please
statuscvo [17]

Answer:

\boxed{\begin{array}{c|c|c|c|c|c|c|c} \bf x &\rm -3 &\rm -2 &\rm -1 &\rm 0 &\rm 1 &\rm 2 &\rm 3 \\\\ \bf y & \rm 0 &\rm -4 &\rm -6&\rm -6 &\rm -4&\rm 2 &\rm 6 \end{array}}

Step-by-step explanation:

A quadratic function is given to us . And we need to fill out the table by using the function . The given function is ,

\rm\implies y = x^2+x - 6

Here we need to substitute the different values of x , to get the different values of y.

<u>Put</u><u> </u><u>x </u><u>=</u><u> </u><u>-</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 3^2-3-6\\

\rm\implies y = 9 - 3 - 6 \\

\rm\implies y = 0

<u>Put </u><u>x </u><u>=</u><u> </u><u>-</u><u>2</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 2^2-2-6\\

\rm\implies y = 4 -2-6\\

\rm\implies y = -4

<u>Put</u><u> </u><u>x </u><u>=</u><u> </u><u>-</u><u>1</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = -1^2-1-6\\

\rm\implies y = 1 -1-6 \\

\rm\implies y = -6

<u>Put </u><u>x </u><u>=</u><u> </u><u>1</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 1^2+1-6\\

\rm\implies y = 2 -6

\rm\implies y = -4

<u>Put </u><u>x </u><u>=</u><u> </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 2^2+2-6\\

\rm\implies y = 4 +2-6\\

\rm\implies y = 2

<u>Put </u><u>x </u><u>=</u><u> </u><u>3 </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 3^2+3-6\\

\rm\implies y = 9 +3-6\\

\rm\implies y = 6

<u>Final </u><u>table</u><u> </u><u>:</u><u>-</u><u> </u>

\boxed{\begin{array}{c|c|c|c|c|c|c|c} \bf x &\rm -3 &\rm -2 &\rm -1 &\rm 0 &\rm 1 &\rm 2 &\rm 3 \\\\ \bf y & \rm 0 &\rm -4 &\rm -6&\rm -6 &\rm -4&\rm 2 &\rm 6 \end{array}}

7 0
3 years ago
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