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Dmitrij [34]
3 years ago
15

3 = 2 + n/4(that's supposed to be a fraction)someone help solve solve for n​

Mathematics
2 answers:
Verizon [17]3 years ago
6 0

Answer:

n = 4

Step-by-step explanation:

take the 2 and subtract it from itself then you gotta do it to the other side so it be 3 -2 and you get 1 then you multiply 4 to n/4 and it crosses it out and what you do to one side you do to the other so you multiply 4 by 1 & get 4

nexus9112 [7]3 years ago
5 0

Answer:

n = 4

Step-by-step explanation:

3 = 2 + n/4

1 = n/4

4 = n

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Austin drove 455 miles in 7 hours.
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Answer:

9 hours

Step-by-step explanation:

455/7 gives you the mph

which is 65

then divide 585 miles by the mph (65)

585/65

= 9 hours

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3 years ago
F= 8/5 (c) + 40 when c=5
Molodets [167]

Answer:

F = 48

Step-by-step explanation:

F= 8/5 (c) + 40

Let c=5

F = 8/5 * 5 +40

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3 years ago
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Its c I've done this before
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3 years ago
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olganol [36]
From the figure, the vertices of the given triangle are
(-3,4), (1,3) and (5,5).

The given transformation is (x,y) -> (x+3, y-2).
Therefore the coordinates of the transformed triangle are
(-3+3,4-2), (1+3,3-2), (5+3,5-2) = (0,2), (4,1), (8,3)

The transformed triangle is shown in red color.

5 0
3 years ago
you drop a ball from a height of 98 feet. at the same time, your friend throws a ball upward. the polynomials represent the heig
Ostrovityanka [42]

a) h_0 -u_y t

b) See interpretation below

Step-by-step explanation:

a)

The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.

Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:

g=32 ft/s^2

in the downward direction.

For the ball dropped from the initial height of h_0 = 98 ft, the height at time t is given by

h(t) = h_0 -\frac{1}{2}gt^2 (1)

The ball which is thrown upward from the ground instead is fired with an initial vertical velocity u_y, and its starting height is zero, so its position at time t is given by

h'(t)=u_y t - \frac{1}{2}gt^2 (2)

Therefore, the polynomial that represents the distance between the two balls is:

h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t

b)

Now we interpret this polynomial, which is:

\Delta h(t) = h_0 -u_y t

which represents the distance between the two balls at time t.

The interpretation of the two terms is the following:

- The constant term, h_0, is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, h_0 = 98 ft

- The coefficient of the linear term, u_y, is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by u_y feet.

5 0
3 years ago
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