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Vitek1552 [10]
3 years ago
13

Evaluate the formula V= Bh/3 for B = 15 in.2 and h = 28 in.

Mathematics
2 answers:
Gelneren [198K]3 years ago
7 0
V=140in.^3 would be the answer
ad-work [718]3 years ago
6 0
V = Bh/3
V = 15 x 28 / 3 
V =  420 / 3
V = 140 in^3
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The slope of the line below is -3. Is the coordinates of the labeled point to find a point - slope equation of the line
puteri [66]
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where y and x stay the same

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and (a, b) (any coordinate on the graph)

based on the point (5, -7)

Y - (-7) = -3 (x - 5)

however, the - cancels out the negative 7 making I positive

Y + 7 = -3(x - 5)












3 0
3 years ago
Find the vertex form of the quadratic function by completing the square <br><br> F(x) = x^2 -24x
musickatia [10]

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Step-by-step explanation:

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3 years ago
What is the slope of the line that passes through the points (4, 7) and (5,8)?
slava [35]

Answer:

1

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8 0
3 years ago
At 3:30 p.m., Berto’s train was 34 miles past the egg farm, traveling at an average speed of 85 miles per hour. At the same time
timurjin [86]

E=egg farm

x=distantce between  the Berto´s trains and the meeting point.

t=time where the trains meet up

⇒110 miles/h                                          ⇒85miles/h

[---------------------------E----------------------]-------------------------------------X

[............12 miles.......][........34 miles.....][------------ x ----------------------]

distance between Berto´s train and Eduardo´s train=

=12 miles+34 miles=46 miles.

S=d/t      ⇒d=S*t

S=seed

d=distance

T=time

Eduardo´s train;

46 miles+x=t*110 miles/h  ⇒x=t*110 miles/h-46 miles  (1)

Berto´s train;

x=t*85 miles/h  (2)

With the equations (1) and (2) we suggest this system equations:

x=t* 110 miles/h-46 miles

x=t*85 miles/h

we solve this system equiations :

t*110 miles/h-46 miles=t*85 miles/h

110t miles/h-85t miles/h=46 miles

25 t miles/h=46 miles

t=46 miles/25 miles/h=1,84 hour  (≈1 hour, 50 minutes, 24 seconds)

x=t*85 miles/h=156,4 miles

Solution: 1,84 hour

3 0
3 years ago
Read 2 more answers
A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe
qaws [65]

The projectile's horizontal and vertical positions at time t are given by

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y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2

where g=9.8\dfrac{\rm m}{\mathrm s^2}. Solve y=0 for the time t it takes for the projectile to reach the ground:

30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s

In this time, the projectile will have traveled horizontally a distance of

x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m

The projectile's horizontal and vertical velocities are given by

v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ

v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about \sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}.

7 0
3 years ago
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