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GaryK [48]
3 years ago
6

A boat's value over time is given as the function f(x) and graphed below. Use A(x) = 400(b)x + 0 as the parent function. Which g

raph shows the boat's value increasing at a rate of 25% per year?

Mathematics
2 answers:
ser-zykov [4K]3 years ago
3 0

Answer:

Please see attached graph

Step-by-step explanation:

We know that the equation tht represents the boats avlue is given by

A(x) = 400(b)^x + 0 = 400(b)^x

A(x) = 400(b)^x

An increment in the rate of 25% a yeard is given by

b = 1.25

A(x) = 400(1.25)^x

Which graph can be seen below

gizmo_the_mogwai [7]3 years ago
3 0

Answer:

It was the first one but included picture just to make sure its the same for everyone. :)

Step-by-step explanation:

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Answer:

0.1m/s²

Step-by-step explanation:

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3 years ago
Omar can jog 3 2/5 miles in 2/3 of an hour. Find his average speed in miles per hour.
Fudgin [204]

Answer:

9.6miles per hour

Step-by-step explanation:

2/3=40min

32/5 /40=0.16

0.16miles per min

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3 0
3 years ago
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A. When x = -1, what is the value of y?
pogonyaev

Answer:

i think we need more context to this answer.  I’d love to help no need to retype. You can just message me for help :)

Step-by-step explanation:

7 0
3 years ago
Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30
Dafna11 [192]
Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}

\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~  \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------

\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o
8 0
3 years ago
I need help with any of these questionss
Iteru [2.4K]

Answer:

sorry hindi ko alam hirapuwu

6 0
3 years ago
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