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Eddi Din [679]
3 years ago
5

Find the constant of proportionality k. Then write an equation for therelationship between xandy(Example 2)2. X8 16y 10 2030 40E

SSENTIAL QUESTION CHECK-IN4. How can you represent a proportional relationship using an equation?20 nit 92432

Mathematics
1 answer:
klio [65]3 years ago
8 0
2) y= x times 5
The constant proportion is k=5

3) y=x times 4
The constant proportion is k=4

4)If total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn.
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What mathematical term describes both 5.3 and 8.2 in the expression 5.3x-8.2y+11.1?
kiruha [24]

Answer:

Step-by-step explanation:

Coefficients

5 0
3 years ago
The volume of the right triangular prism is 91.8ft. The height of the prism is 10.8ft. What is the area of each base? Show your
kaheart [24]

Answer:

8.5 ft.

Step-by-step explanation:

We'll use this right triangular prism volume formula: base area x height = volume

base area = ?

length = 10.8

volume = 91.8

base area x 10.8 = 91.8

Divide both sides by 10.8

base area x 10.8/10.8 = 91.8/10.8

base area = 8.5

4 0
3 years ago
2. Four friends share 3 apples equally.
Slav-nsk [51]

Answer:

1/3

Step-by-step explanation:

Each friend gets 1/3 of an apple

5 0
3 years ago
5x7=<br> 5x7=5 x (_ + _)<br> 5x7= (5x _) + (5x_)<br> 5x7= _ + _<br> 5x7=
eimsori [14]
5x7=35
5x7=5x(5+2)
5x7=(5x3)+(5x4)
5x7=20+15
5x7=35

This is how I would do it :)
7 0
3 years ago
Read 2 more answers
PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS
Elina [12.6K]

You want to end up with A\sin(\omega t+\phi). Expand this using the angle sum identity for sine:

A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi

We want this to line up with 2\sin(4\pi t)+5\cos(4\pi t). Right away, we know \omega=4\pi.

We also need to have

\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}

Recall that \sin^2x+\cos^2x=1 for all x; this means

(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}

Then

\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)

So we end up with

2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)

8 0
3 years ago
Read 2 more answers
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