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3 years ago

Plz help ill give u brainlist

Mathematics

1 answer:

Basile [38]3 years ago

7 0

2x + 40
Divide both sides by 2
You get x + 20

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Find the distance C by using Phytogreon Theorem & distance formula B (-1,5 . -4,5)

kirza4 [7]

Answer:

3 lol

Step-by-step explanation:

5 0

2 years ago

Julia is standing 2 feet away from a lamppost. If she casts a shadow of 5 feet and the light makes a 20° angle relative to the g

givi [52]

Answer:

2.5477 feet

Step-by-step explanation:

Refer the image attached to understand my solution.

BC = height of lamppost.

DE = Julia

AD = shadow of Julia

BD = 2 feet. AD = 5 feet

BA = BD + AD

= 2 + 5 = 7 feet

In ΔABC

tan 20° = $\frac{BC}{BA} = \frac{BC}{7}$

BC = 7 * tan 20°

BC = 2.5477 feet

SO the height of lamppost = 2.5477 feet

5 0

2 years ago

On a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 195 feet at an angl

worty [1.4K]

In this problem, we can imagine that all the points connect to form a triangle. The three point or vertices are located on the pitcher mount, the home plate and where the outfielder catches the ball. So in this case we are given two sides of the triangle and the angle in between the two sides.

With the following conditions, we can use the cosine law to solve for the unknown 3rd side. The formula is:

c^2 = a^2 + b^2 – 2 a b cos θ

Where,

a = 60.5 ft

b = 195 ft

θ = 32°

Substituting the given values:

c^2 = (60.5)^2 + (195)^2 – 2 (60.5) (195) cos 32

c^2 = 3660.25 + 38025 – 20009.7

c^2 = 21,675.56

c = 147.23 ft

Therefore the outfielder throws the ball at a distance of 147.23 ft towards the home plate.

8 0

2 years ago

Agent Bond is standing on a bridge, 13.5 m above the road below, and his pursuers are getting too close for comfort. He spots a

Scrat [10]

Answer:

Step-by-step explanation:

Eek! Let's give this a go. Things we know:

acceleration of Bond in free fall is -9.8 m/s/s

velocity of the truck is 25 m/s

displacement Bond will travel when he jumps is -10 m

What we are looking for is the time it will take him to hit the top of the truck, knowing that the truck can travel from one pole to the next in 1 second.

Our displacement equation is

Δx = v₀t + 1/2at²

Filling in we have

$-10=25t+\frac{1}{2}(-9.8)t^2$

Simplifying we get

$-10=25t-4.9t^2$

This is a quadratic that needs to be solved however you personally solve quadratics. When you do that, you find that the times it will take Bond to drop that displacement is either -.37 seconds or 5.47 seconds. Many things in physics can be negative, like velocity and acceleration, but time NEVER will be. So it takes Bond 5.5 seconds to drop to the roof of the moving truck. That means that he needs to jump when the truck is between the 5th and the 6th poles away from him.

Good luck with this!

Cheers!

6 0

3 years ago

What is the LCM of 8 and 10

iragen [17]

40

8x5=40
10x4=40

Hope this helps

3 0

2 years ago