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Ludmilka [50]
3 years ago
6

When adding with exponents, what do you to the exponents?

Mathematics
1 answer:
Svet_ta [14]3 years ago
8 0

To add with powers, both the variables and the exponents of the variables must be the same. You perform the required operations on the coefficients, leaving the variable and exponent as they are. When adding with powers, the terms that combine always have exactly the same variables with exactly the same powers

http://www.wikihow.com/Add-Exponents#Adding_Variables_with_Exponents_sub

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. What are the equation and slope of the line that is parallel to the y-axis and goes through the point (1,-2)?​
Brrunno [24]
The answer:

x = 1

Explanation:

If the line is parallel to the y-axis it has no slope, so you would only write down the x-intercept.
8 0
3 years ago
Identify the perimeter and area of a square with diagonal length 11in. Give your answer in simplest radical form. HELP PLEASE!!
Nataliya [291]

Answer:

  • Perimeter = 22*sqrt(2)
  • Area = 60.5 inches
  • D

Step-by-step explanation:

Remark

You need 2 facts.

  1. A square has 4 equal sides.
  2. It contains (by definition) 1 right angle but since we are not including and statement about parallel sides, it needs 4 right angles.

That means you can use the Pythagorean Theorem.

If one side of a square is a then the 1 after it is a as well.

Formula

  • a^2 + a^2 = c^2
  • 2a^2 = c^2

Givens

  • c = 11

Solution

  • 2a^2 = 11^2
  • 2a^2 = 121                    Divide by 2
  • a^2 = 121/2                  Take the square root of both sides
  • sqrt(a^2) = sqr(121/2)    
  • a = 11/sqrt(2)                Rationalize the denominator
  • a = 11 * sqrt(2)/[sqrt(2) * sqrt(2)]
  • a = 11 * sqrt(2) / 2

<em><u>Perimeter</u></em>

P = 4s

  • P = 4*11*sqrt(2)/2
  • P = 44*sqrt(2)/2
  • P = 22*sqrt(2)

You don't need the area. The answer is D

<em><u>Area</u></em>

  • Area = s^2
  • Area = (11*sqrt(2)/2 ) ^2
  • Area = 121 * 2 / 4
  • Area = 60.5


8 0
3 years ago
PLEASE HELP IVE ASKED THIS QUESTION 3 TIMES PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!​
worty [1.4K]

THEOREM:

  • h² = p² + b² where h is hypotenuse, b is base and p is perpendicular.

ANSWER:

[3] By pythagorean theorem,

  • x² = 14² + 9²
  • x² = 196 + 81
  • x² = 277
  • x = √277
  • x = 16.64 rounded.

[4] By pythagorean theorem,

  • x² = 32² + 24²
  • x² = 1024 + 576
  • x² = 1600
  • x = √1600
  • x = 40.

[5] By pythagorean theorem,

  • (2x)² = 21² – 12.6²
  • 4x² = 441 – 158.76
  • 4x² = 284.24
  • x² = 284.24/4 = 70.56
  • x = √70.56
  • x = 8.4

[6] By tangent property,

  • 7x – 29 = 2x + 16
  • 7x – 2x = 16 + 29
  • 5x = 45
  • x = 9.

So, WX = 7(9) – 29 = 63 – 29

  • WX = 34
4 0
3 years ago
A pumpkin is launched from the top of a 20 foot tall platform at an initial velocity of 84 feet per second. The height, h, of th
Gennadij [26K]
<h2>Explanation:</h2>

In this problem we have a pumpkin is launched from the top of a 20 foot tall platform at an initial velocity of 84 feet per second. So the height, h, of the pumpkin at time t seconds after the launch can be modeled by the equation:

h(t) = -16t^2 + 84t + 20

So this is the equation of a parabola. The maximum of this parabola occurs at its vertex. So let's find this vertex:

\text{The x-value of the vertex is}: \\ \\ t=-\frac{b}{2a} \\ \\ \\ Where: \\ \\ a=-16 \\ \\ b=84 \\ \\ c=20 \\ \\ \\ t=-\frac{84}{2(-16)} \\ \\ t=-\frac{84}{-32} \\ \\ t=\frac{21}{8}=2.625 \\ \\ h(2.625)=-16(2.625)^2+84(2.625)+20 \\ \\ h(2.625)=-110.25+220.5+20 \\ \\ h(2.625)=130.25m

Finally, the maximum occurs at time 2.625 seconds when the height is 130.25m

8 0
4 years ago
Im confused and i need help please help
denis23 [38]

Answer:

C

Step-by-step explanation:

8 0
3 years ago
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