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3 years ago

Pls help meeeeeeeeeeeeeee

Mathematics

2 answers:

horsena [70]3 years ago

8 0

It is c, 73. all you do is plug 4 in for x then multiply and add

MArishka [77]3 years ago

5 0

Answer:

Step-by-step explanation:

13+15(4) = 73

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How much time has passed? Start time: 1:50 p.m. End time: 4:40 p.m.

Romashka-Z-Leto [24]

Answer:

2 hours and 10 minutes

Step-by-step explanation:

Sorry it took a while...

Hope this helped! :)

8 0

3 years ago

So I didn’t get a proper answer my last question and I need this done nowwwwwww it’s pre-algebra!

ddd [48]

Answer:

1. Anything to do with the same slope but different a B (Ex. y=1/3x+1 and y=1/3x-8)

2. No solutions

Step-by-step explanation:

1. Because the two lines are parallel, when no intersection between two lines happens, there will need to be a change in the B but the same slope. Remember the formula for an equation is y=mx+b. The m stands for the slope. With parallel lines, slopes are the same. If lines were to never intersect, you would have to change the b to a different number so that both lines could keep going at the same time, but never intersect.

2. Remember, solutions only happen when the two lines intersect. Infinite solutions happen when the lines are the same. One solution happens when two lines only have one point where they intersect. No solutions happen when the two lines don't intersect at all. Based off the graph given, there are no intersections so, there will be no solutions.

5 0

2 years ago

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....

$0=-16t^2+500t+20\\ \\ -4(4t^2-125t-5)=0\\ \\ 4t^2-125t-5=0$

Using quadratic formula, we will get.....

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(-5)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{15705}}{8}\\ \\ t=31.2899...\\ and\\ t= -0.0399...$

(Negative value is ignored as time can't be in negative)

So, the rocket will remain in the air for 31.2899... seconds.

5 0

4 years ago

A new airport is in its planning phase. One particular challenge is that there are several tall buildings in the vicinity. In pa

mixer [17]

Answer:

Could an airplane take off and clear this tall building? YES

Step-by-step explanation:

Trigonometry

The building and the ground form a right (90°) angle. The path of the airplane (assumed a straight line) completes the right triangle.

The takeoff angle of the plane θ=15° has the height of the building (450 feet) as the opposite side and the horizontal distance from the end of the runway (2500 feet) as the adjacent side.

The tangent of θ is defined as the following ratio:

$\displaystyle \tan\theta=\frac{\text{opposite leg}}{\text{adjacent leg}}$

$\displaystyle \tan \theta=\frac{450}{2500}$

$\displaystyle \tan \theta=0.18$

Calculating the inverse tangent function:

$\theta=\arctan 0.18$

$\theta\approx 10^\circ$

This means the angle needed to clear the tall building is about 10° and it's within the maximum airplane's takeoff angle.

Could an airplane take off and clear this tall building? YES

8 0

3 years ago

One batch of muffins require 1/4 cup of sugar. Kelly wants to make 3/2 batch of muffins. Will she use more than, less than, or a

Alla [95]

She will use more than 1/4 cup of sugar to make 1.5 batches of muffins. If one batch of muffins requires 1/4 cup of sugar, you can think of the amount of sugar needed being 1/4 cup per batch. That being said multiply 1.5 batches by 1/4 cup per batch to find cups of sugar.
(1.5 batches)x((1/4 cups)/(1 batch))=0.375 cups of sugar.

You can also think of it as if one batch takes 1/4 cup of sugar, half of a batch would take 1/8 cup of sugar. Since 1.5 batches is just 1 batch+1/2 batch, you can say the amount of sugar needed is 1/4 cup+1/8 cup which equals 3/8 cup sugar.

I hope this helps you.

5 0

4 years ago