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Sunny_sXe [5.5K]
3 years ago
13

(1 + 2) x 2 it equals 6

Mathematics
1 answer:
torisob [31]3 years ago
4 0

Answer:

yes you have to use PEMDAS

Step-by-step explanation:

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Which one should I choose
erik [133]

Answer: 9

Step-by-step explanation:

Since x=2 that means there should be 2 3s being multiplied since it’s 3 to the second power. So 3x3 which equals 9. If this answer helped please mark it brainliest!

3 0
3 years ago
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NEED HELP 15 POINTS!!!!1
xz_007 [3.2K]

Answer: (D) ∠2 is supplementary to ∠8

<u>Step-by-step explanation:</u>

∠2 is congruent to:

  • ∠3   <em>vertical angles</em>
  • ∠6    <em>corresponding angles</em>
  • ∠7   <em>alternate exterior angles</em>

Therefore, it is supplementary to the remaining angles:

  • ∠1
  • ∠4
  • ∠5
  • ∠8

The only multiple choice option that satisfies the above statements is the last one

5 0
3 years ago
What is the estamate of 1374 times 6
Gemiola [76]
Hey!~

To make an "estimate", you can round 1374 to the nearest hundreds place, making it 1400. Next, all you need to do is do 1400*6, which is simply 8400.

Hope this helps and have a great day!

-Lindsey
3 0
3 years ago
Patrick wants to prepare a salad. He needs 3 cups of cooked macaroni, 3 cups of sliced oranges, 2 cups of sliced apple, 1 cup of
Gnesinka [82]

Answer:

s

Step-by-step explanation:

4 0
3 years ago
Two buildings on opposites sides of a highway are 3x^3- x^2 + 7x +100 feet apart. One building is 2x^2 + 7x feet from the highwa
iVinArrow [24]

Given:

Distance between two buildings = 3x^3- x^2 + 7x +100 feet apart.

Distance between highway and one building = 2x^2 + 7x feet.

Distance between highway and second building = x^3 + 2x^2 - 18 feet.

To find:

The standard form of the polynomial representing the width of the highway between the two building.

Solution:

We know that,

Width of the highway = Distance between two buildings - Distance of both buildings from highway.

Using the above formula, we get the polynomial for width (W) of the highway.

W=3x^3- x^2 + 7x +100-(2x^2 + 7x)-(x^3 + 2x^2 - 18)

W=3x^3- x^2 + 7x +100-2x^2-7x-x^3 -2x^2+18

Combining like terms, we get

W=(3x^3-x^3)+(- x^2 -2x^2-2x^2)+ (7x -7x)+(100 +18)

W=2x^3-5x^2+0+118

W=2x^3-5x^2+118

Therefore, the width point highway is 2x^3-5x^2+118.

8 0
3 years ago
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