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3 years ago

I’ll mark brainliest

Mathematics

1 answer:

Naddik [55]3 years ago

4 0

Answer:

Step-by-step explanation:

ok so pretend the 2.00 was just 2 so whats 8 divided by 2? 4 right so if we test that witht he other problems its the same so its A

Hope this helps!!!! :D

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There are 5000 sweets 24% blue 5% green 17% red rest are orange find percentage of orange

ss7ja [257]

Answer: 54%

Step-by-step explanation:

24+17+5=46

100-46=54

5 0

3 years ago

Can you help me please

Olin [163]

Answer:

(2,1)

Step-by-step explanation:

3 0

3 years ago

I need help!! on 9 and 10 plzz help

Alexxandr [17]

9. 1/3+(x+20)+(x-10)+40=360 10. 1/2x+1/2x+(x-15)+(x-25)+100=540

1/3+2x+50=360 4x+60=540

(3/1) 1/3+2x=300(3/1) 4x/4=480/4

2x=300 X=120

————

W=150

6 0

4 years ago

A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of

Ivenika [448]

The equation for the height of the rocket at time t given

$h= -16t^2+192t$

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

$h= -16t^2+192t$

$560 = -16t^2+192t$

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

$560 = -16(t^2 - 12t)$

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

$560/-16 = -16(t^2-12t)/-16$

$-35 = t^2 -12t$

Now we will move -35 to the righ side by adding 35 to both sides.

$-35+35 = t^2-12t+35$

$0 = t^2 -12t+35$

$t^2-12t+35 = 0$

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

$t^2-12t+35 =0$

$(t-5)(t-7) =0$

So by using zero product property we will get

$t-5 =0$

$t-5+5 = 0+5$

$t=5$

Also $t-7 =0$

$t-7+7 = 0+7$

$t=7$

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

$h= -16t^2+192t$

$0= -16t^2 + 192 t$

$0 = -16(t^2-12t)$

$-16(t^2-12t) = 0$

We will move -16 to the other side by dividing it to both sides.

$-16(t^2-12t)/-16 = 0/-16$

$t^2-12t = 0$

We will take out the common factor t from the left side. By taking out t we will get,

$t(t-12) = 0$

We will use zero product property now. By using that we will get,

$t = 0$

ans also $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

6 0

3 years ago

Sin(2x+30°)=1 <br> Tìm x <br> Cảm ơn vì giúp đỡ

Pie

Answer:

Step-by-step explanation:

This question is asking us to find where sin(2x + 30) has a sin of 1. If you look at the unit circle, 90 degrees has a sin of 1. Mathematically, it will be solved like this (begin by taking the inverse sin of both sides):

$sin^{-1}[sin(2x+30)]=sin^{-1}(1)$

On the left, the inverse sin "undoes" or cancels the sin, leaving us with

2x + 30 = sin⁻¹(1)

The right side is asking us what angle has a sin of 1, which is 90. Sub that into the right side:

2x + 30 = 90 and

2x = 60 so

x = 30

You're welcome!

4 0

3 years ago