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xz_007 [3.2K]
3 years ago
15

How do we get 144 here?

Mathematics
1 answer:
Brums [2.3K]3 years ago
6 0

Answer:

the correct answer is supposed to be 120, your almost there :)

Step-by-step explanation:

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Solve pls<br> Ill give brainliest
oksian1 [2.3K]

Answer:

A

Step-by-step explanation:

it cannot be B nor C there for we only have A and D tho D says that there are 7 players there might be more in/out not counting in, there for the answer is A.

7 0
3 years ago
Choose the point slope form of the equation below that represents the line that passes through the points (-3, 2) and (2, 1)
mihalych1998 [28]

Answer:

\large\boxed{y-1=-\dfrac{1}{5}(x-2)}

Step-by-step explanation:

The point-slope form of a line:

y-y_1=m(x-x_1)

m - slope

(x₁, y₁) - point

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (-3, 2) and (2, 1). Substitute:

m=\dfrac{1-2}{2-(-3)}=\dfrac{-1}{5}=-\dfrac{1}{5}

y-1=-\dfrac{1}{5}(x-2)

4 0
3 years ago
Write a function describing the relationship of the given variables.
Oliga [24]

Answer:

The function describing the relationship of V and t is V = 3t²

Step-by-step explanation:

* Lets explain the meaning of direct variation

- The direct variation is a mathematical relationship between two

  variables that can be expressed by an equation in which one

  variable is equal to a constant times the other

- If Y is in direct variation with x (y ∝ x), then y = kx, where k is the

 constant of variation

* Now lets solve the problem

# V is varies directly with the square  of t

- Change the statement above to a mathematical relation

∴ V ∝ t²

- Chang the relation to a function by using a constant k

∴ V = kt²

- To find the value of the constant of variation k substitute V and t

 by the given values

∵ t = 6 when V = 108

∵ V = kt²

∴ 108 = k(6)² ⇒ simplify the power 2

∴ 108 = 36k ⇒ divide both sides by 36 to find the value of k

∴ 3 = k

- The value of the constant of variation is 3

∴ The function describing the relationship of V and t is V = 3t²

3 0
3 years ago
During the first stages of an epidemic, the number of sick people increases exponentially with time. Suppose that at = 0 days th
nydimaria [60]
T = days passed

r = rate of growth

by 0 day, or t = 0, there are 2 folks sick,

\bf \qquad \textit{Amount for Exponential Growth}&#10;\\\\&#10;A=P(1 + r)^t\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &2\\&#10;P=\textit{initial amount}\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=\textit{elapsed time}\to &0\\&#10;\end{cases}&#10;\\\\\\&#10;2=P(1+r)^0\implies 2=P\cdot 1\implies 2=P\qquad \boxed{A=2(1+r)^t}

 by the third day, t = 3, there are 40 folks sick,

\bf \qquad \textit{Amount for Exponential Growth}&#10;\\\\&#10;A=P(1 + r)^t\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &40\\&#10;P=\textit{initial amount}\to &2\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=\textit{elapsed time}\to &3\\&#10;\end{cases}&#10;\\\\\\&#10;40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r&#10;\\\\\\&#10;\sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}

how many folks are there sick by t = 6?   \bf \stackrel{that~many}{A=2(2.7)^6}
8 0
3 years ago
A jewelry box containes 3 rings, 7 necklaces, 4 earrings, 5 braceletes and 1 tiara. If 1 piece of jewlery is selected at radom,
Brums [2.3K]

Answer:

C and D

Step-by-step explanation:

Hope this helps!

6 0
3 years ago
Read 2 more answers
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