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3 years ago

I NEED HELP! simplify this expression

Mathematics

1 answer:

ioda3 years ago

7 0

Answer:

Step-by-step explanation:

36-1*16+5 (PEMDAS, exponents first)

36-16+5 (PEMDAS, multiplication)

20+5 (PEMDAS, add and subtract left to right)

25 (PEMDAS, add and subtract left to right)

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What is the interquartile range of the data set?

Andrej [43]

Answer:

Should be 10 if there is nothing for th stem value of 3.

Step-by-step explanation:

The stem is the tens place, and the leaf is the ones. You would get 28 40 43 43 45 50 50 for your numbers. If you take Q3-Q1, you get your answer, 10.

7 0

2 years ago

FIND THE AREA AND PERIMETER PLEASE

baherus [9]

Answer:

Area = 70 Perimeter = 50

Step-by-step explanation:

I will begin with area. I can find the area of each figure, then add the areas.

3 • 18 = 54

4 • 4 = 16

Area = 70

For perimeter, I will add the given lengths.

3 + 18 + 7 + 4 + 4 + 14

Perimeter = 50

8 0

3 years ago

Read 2 more answers

(-2,3) and slope m=9

klemol [59]

Answer:

Step-by-step explanation:

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2 years ago

The pie recipe calls for twice as many peaches as nectarines. If it takes a total of 168 pieces of fruit for all the pies, how m

Inessa05 [86]

84 since 84 times 2 equals 168

4 0

3 years ago

Solving for matrices

muminat

Answer:

Step-by-step explanation:

The augmented matrix for the system of three equaitons is

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)$

Multiply the first row by 5, the second row by -3 and add these two rows:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)$

Subtract the third row from the second:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)$

Divide the third row by 6:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)$

Now multiply the third equation by 26 and add it to the second row:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)$

You get the system of three equations:

$\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.$

From the third equation

$z=\dfrac{90}{45}=2.$

Substitute z=2 into the second equation:

$-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.$

Now substitute z=2 and y=5 into the first equation:

$3x-4\cdot 5-5\cdot 2=-27\\ \\3x-20-10=-27\\ \\3x-30=-27\\ \\3x=-27+30=3\\ \\x=1.$

The solution is (1,5,2)

4 0

2 years ago