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3 years ago

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Write the following repeated multiplication using exponents. 66666 (**Please use the ^ sign to show powers; for example, 5*5

=5^2.**) *

1 answer:

kondaur [170]3 years ago

5 0

Answer:

6^5

Step-by-step explanation:

6×6×6×6×6=7776

So $6^{5}$ =7776

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THERE IS MORE THAN ONE :(

ch4aika [34]

Answer:

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Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Work out the perimeter of this semicircle with radius 3cm. Give your answer in terms of π

kondaur [170]

Step-by-step explanation:

<u>Given </u><u>:</u><u>-</u><u> </u>

A semicircle is given with radius of 3cm .

And we need to find the perimeter of the semicircle . We know the perimeter as ,

$:\implies$ Perimeter = πr

$:\implies$ Perimeter = 3cm × π

$:\implies$ Perimeter = 3π cm .

Now there is the diameter of circle is 6cm . Hence totally the perimeter is ,

$:\implies$ Perimeter = 3π + 6 cm

We were supposed to give answers in terms of π .

<h3>Hence the perimeter is ( 3π + 6 ) cm</h3>

5 0

2 years ago

agasfer [191]

Ok im not sure the answer yet so ima work on it at the same time while im explaining it. (The answer will probably be near the end)

We can use the elimination method to eliminate y out. To do that we multiply the first equation by 3.

6x+3y=-12

Now just subtract it from the other equation.

6x+3y=-12
5x+3y=-6

***x=-6***

Usually after doing the elimination method you will have to solve for x but in this case its already solved for you. If you want to find y now you just take the first equation and fill x with -6 and solve for y.

2(-6)+y=-4
-12+y=-4
y=8

Brainliest my answer if it helps you out?

4 0

3 years ago

Read 2 more answers

1.) What is the equation of the path of firework #1? Write your equation in general form.

valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in $y=ax^2+bx+c$

it's easier to find vertex form first then expand to get general form

vertex form is $y=a(x-h)^2+k$ where the vertex is (h,k) and a is a constant

firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

$h_1=a(t-10)^2+50$

to find the value of $a$ , subsitute another point

(0,0)

$0=a(0-10)^2+50)$

$0=100a+50$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_1=\frac{-1}{2}(t-10)^2+50$

expand to get general form

$h_1=\frac{-1}{2}(t^2-20t+100)+50$

$h_1=\frac{-1}{2}t^2+10t-50+50$

$h_1=\frac{-1}{2}t^2+10t$

2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is $h_2=a(t-10)^2+72$

find $a$

use another point

(0,22)

$22=a(0-10)^2+72$

$22=100a+72$

$-50=100a$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_2=\frac{-1}{2}(t-10)^2+72$

3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50

domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

3 0

3 years ago

*Spam answers will not be tolerated*

V125BC [204]

Step-by-step explanation:

C = -11

A = -13

V = 9

E = -9

If you are adding them: -24

Hope this helps!

8 0

2 years ago