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kow [346]
3 years ago
5

Help me please 123432323

Mathematics
2 answers:
suter [353]3 years ago
8 0
About .46?????????????????????
gladu [14]3 years ago
4 0

Answer: About 0.46

Step-by-step explanation:

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The standard diameter of a penny is 0.75 in. and the standard
Blababa [14]

Answer: .045

Step-by-step explanation:.75=.750 so, 750-705=.045

Hope this helps!

6 0
3 years ago
QUICK !<br> What is the image point of (5,-5) after a translation right 2 units and up 3 units ?
faust18 [17]

Answer:

(7,-2)

Step-by-step explanation:

First, you move to the right of y-axis like 5 to 7,

then you movie to the up of x-axis like -5 to -2.

So the answer was (7,-2)

5 0
3 years ago
Jeremy ate 5 pieces of shrimp, Julie ate 7 pieces. What fraction of the total did Jeremy eat?
matrenka [14]

Answer:

\frac{5}{12}

Step-by-step explanation:

The total amount of shrimp that was eaten was 12 pieces of shrimp.

To make the fraction you will simply put the amount of shrimp Jeremy ate over the total number of shrimp eaten by the pair.

Therefore, the fraction would look like this:

\frac{5}{12}

The decimal form of this would be 0.4167. Which also means that Jeremy ate 41.67% of the total amount of shrimp eaten.

Hope this helps!!

4 0
2 years ago
What is the value of x in the eqation 5x+3=4x
Reika [66]

Answer :x=-3

Step-by-step explanation:Solve for  x by simplifying bothh sides of the equation then isoolating the variable

7 0
2 years ago
Read 2 more answers
A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s
mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, \theta=25°

Initial speed is, u=15\ m/s

Time of flight is, t=3.0\ s

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = -h (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, g=-9.8\ m/s^2

Now, using the following equation of motion;

-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, v_y=0\ m/s

Now, using the following equation of motion, we have:

v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m

Therefore, the maximum height of the brick is 2.05 m.

5 0
3 years ago
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