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KiRa [710]
3 years ago
8

What is z to the 5th times z to the 6th

Mathematics
2 answers:
Troyanec [42]3 years ago
7 0

Z^5xZ^6  you have to multiply and then keep 1 Z

Z^30

nadezda [96]3 years ago
5 0
Z to 5z to z to 6z
that's what I guess is the answer
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A _______ is a set of characters that uses the same typeface.
eimsori [14]
A font is a set of characters that uses the same typeface.It is a specific typeface of a fix size and style. An example is, a font may be Arial 12 pt bold, while another font may be Times New Roman 14 pt italic. According to Merriam dictionary, font is an assortment or set of type or charactersall of one style and sometimes one size.
4 0
3 years ago
1. What is the perimeter of the figure below. Simplify.<br> 4x^3-5x 2x^3+6x 6x^3+x
____ [38]

Answer:

12x^3+2x

Step-by-step explanation:

The perimeter of a polygon is equal to the sum of its sides.

The sides in the triangle shown have lengths 2x^3+6x, 4x^3-5x, and 6x^3+x in no specific order.

Since its perimeter is given by the sum of its sides, the perimeter is equal to:

2x^3+6x+4x^3-5x+6x^3+x

Combine like terms:

\boxed{12x^3+2x}

6 0
3 years ago
An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0
3 years ago
BRAINLIEST IF CORRECT
lbvjy [14]

Answer:

Step-by-step explanation:

the first is 1/2

second is 1/52

third is 1/4

forth is 1/8

i think thats right any way :)

7 0
3 years ago
Read 2 more answers
Which equations are true for x = -2 and x = 2? Select two options
Flauer [41]

Answer:

A and C

Step-by-step explanation:

solving the equations

A

x² - 4 = 0 ( add 4 to both sides )

x² = 4 ( take square root of both sides )

x = ± \sqrt{4} = ± 2 ← required solution

B

x² = - 4 ← has no real solutions

C

4x² = 16 ( divide both sides by 4 )

x² = 4 ( take square root of both sides )

x = ± \sqrt{4} = ± 2 ← required solution

D

2(x - 2)² = 0 , then

x- 2 = 0 ( add 2 to both sides )

x = 2 ← not the required solution

5 0
2 years ago
Read 2 more answers
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