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Aleksandr-060686 [28]
3 years ago
11

What is another way to write the expression ​ p⋅(10−2) ​ ?

Mathematics
1 answer:
Tomtit [17]3 years ago
6 0

Answer: A

Step-by-step explanation:

I think it is A because p*(10-2). Can = (p*10)-(p*2)

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Solve arctan(-sqrt(3))
tatyana61 [14]
We are given with the expression arctan(-sqrt(3)) and asked to evaluate it. In this case, we can use a calculator or the rule of common triangles to answer this question. the value of <span>arctan(-sqrt(3)) is -60. Since negative tan is found in 2nd and 4th quadrant, the angles are 180-60 or 120 degrees and 360-60 or 300 degrees.</span>
6 0
3 years ago
I need question 3 only
dsp73

Answer:

  • table: 14, 16, 18
  • equation: P = 2n +12

Step-by-step explanation:

Perimeter values will be ...

rectangles . . . perimeter

  1 . . . 14

  2 . . . 16

  3 . . . 18

__

The perimeter of a figure is twice the sum of the length and width. Here, the length is a constant 6. The width is n, the number of rectangles. So, the perimeter is ...

  P = 2(6 +n) = 12 +2n

Your equation is ...

  P = 2n +12 . . . . . . . . perimeter P of figure with n rectangles.

_____

<em>Additional comment</em>

You may be expected to write the equation using y and x for the perimeter and the number of rectangles. That would be ...

  y = 2x +12 . . . . . . . . . perimeter y of figure with x rectangles

4 0
3 years ago
3y + x = 12<br> What is the value of x when y = 3?
Andre45 [30]

Step-by-step explanation:

3y + x = 12

When y = 3, we have 3(3) + x = 12.

=> 9 + x = 12, x = 3.

6 0
3 years ago
Which expression is equivalent to the area of square A, in square inches?
mezya [45]

Answer:

the required expression equivalent to the area of the square A in inches is (10² + 24²).

Step-by-step explanation:

7 0
2 years ago
Before sending track and field athletes to the Olympics, the U.S. Holds a qualifying meet. The box plots below show the distance
sleet_krkn [62]

Question:

The options are;

A. The distances in the Olympic final were farther on average.

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

C. The distances in the Olympic final were all greater than the US qualifier distances

D. none of the above

Answer:

The correct option is;

A. The distances in the Olympic final were farther on average.

Step-by-step explanation:

From the options given, we have

A. The distances in the Olympic final were farther on average.

This is true as the sum of the 5 points divided by 5 is more in the Olympic final

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

This is not correct as the difference between the upper and lower quartile in the Olympic final is lesser than in the qualifier

C. The distances in the Olympic final were all greater than the US qualifier distances

This is not correct as the max of the qualifier is more than the lower quartile in the Olympic final

D. none of the above

We have seen a possible correct option in option A

7 0
3 years ago
Read 2 more answers
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