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vichka [17]
2 years ago
15

The perimiter of a rectangle is 128 inches. The width of the rectangle is 4y inches. Find the length of the rectangle in terms o

f y.
PLEASE HELP!
I will give brainiest if it is correct
Mathematics
1 answer:
disa [49]2 years ago
4 0

Answer:

The perimeter of a rectangle is 2W + 2L.

P = 2W + 2L

128 = 2(4y) + 2L = 8y + 2L

2L = 128 - 8y

L = 64 - 4y (this is the answer

Step-by-step explanation:

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sarah spent a total of 10 on oranges and apples at the supet market. if she spent 3 dollars less for oranges than she did on app
beks73 [17]

Let's write 2 equations from the two statements given.

<em>Sarah spent 10 dollars on both oranges and apples</em>

<em />

Let the price of oranges be "x" and price of apples be "y", thus we can write:

x+y=10

Oranges cost 3 less than apples, thus we can say:

y-3=x

We can substitute this into the first equation and solve for y:

\begin{gathered} x+y=10 \\ y-3+y=10 \\ 2y=10+3 \\ 2y=13 \\ y=\frac{13}{2} \\ y=6.5 \end{gathered}

Thus, let's solve for x now,

\begin{gathered} x=y-3 \\ x=6.5-3 \\ x=3.5 \end{gathered}

We want the price of oranges (x), thus,

<em>Price of Oranges = $3.50</em>

3 0
10 months ago
What are the mensure of 1 angle and 2 angle? Show ur work or explain
STatiana [176]

Answer:

angle 1 = 105º

angle 2  = 75º

Step-by-step explanation:

angle 2  = 75º

corresponding angle to the 75º angle

angle 1 = 105º

it's supplementary to angle 2

180 - 75 = 105

6 0
2 years ago
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Papessa [141]

Answer:

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Step-by-step explanation:

6 0
2 years ago
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Which are partial products for 68 × 43?
Ksivusya [100]
D.600 TINEE 40 =24,000
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2 years ago
A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739
OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

  1. For a constant acceleration: v_{f}=v_{0}+at, where  v_{f} is the final velocity in a direction after the acceleration is applied, v_{0} is the initial velocity in that direction  before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
  2. <em>Then for the x direction</em> it is known that the initial velocity is v_{0x} = 5320 m/s, the acceleration (the applied by the engine) in x direction is a_{x} 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the  is 739 s. Then: v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}
  3. In the same fashion, <em>for the y direction</em>, the initial velocity is  v_{0y} = 0 m/s, the acceleration in y direction is a_{y} 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}
8 0
2 years ago
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