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GREYUIT [131]
3 years ago
14

Travis wants to collect 20% more than Jessica. Robin wants to collect 35% more than Travis. If Travis collects $43, how much was

collected in all?
Mathematics
1 answer:
Cerrena [4.2K]3 years ago
4 0
Travis collected 43. To find what Jessica collected, you do 43 - (43*0.20) = 34.4 or $34.40. To find what Robin collected, you do 43 + (43*0.35) = 58.05 or $58.05.

Add these together: 43.00 + 34.40 + 58.05 = 135.45.

$135.45 were collected in total assuming everyone collected exactly the percentage they 'wanted' to. 
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Use the cubic model y = 5a3 - 2a2 + a - 45 to find the value of y when x = 4.
Naddik [55]

Assuming 5a3 equals 5a^3, and the same for everything else, you would plug in 4 for a, so 5(4)^3 - 2(4)^2 + 4 - 45, which equals 247.

3 0
3 years ago
PLEASE HELP 100 POINTS!!!!!!
horrorfan [7]

Answer:

A)  See attached for graph.

B)  (-3, 0)  (0, 0)  (18, 0)

C)   (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.

Given piecewise function:

g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad  \textsf{if }x\geq 3\end{cases}

Therefore, the function has two definitions:

  • g(x)=x^3-9x \quad \textsf{when x is less than 3}
  • g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}

<h3><u>Part A</u></h3>

When <u>graphing</u> piecewise functions:

  • Use an open circle where the value of x is <u>not included</u> in the interval.
  • Use a closed circle where the value of x is <u>included</u> in the interval.
  • Use an arrow to show that the function <u>continues indefinitely</u>.

<u>First piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=(3)^3-9(3)=0 \implies (3,0)

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

<u>Second piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

<h3><u>Part B</u></h3>

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the <u>first piece</u> of the function to zero and solve for x:

\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the <u>second piece</u> to zero and solve for x:

\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}

\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b

\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

<h3><u>Part C</u></h3>

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation:  (-3, 0) ∪ [3, 18)

Learn more about piecewise functions here:

brainly.com/question/11562909

3 0
2 years ago
Need help finding a constant k at which the function is continuous everywhere.
saveliy_v [14]

Answer:

k = -6/35

Step-by-step explanation:

To make the function continuous

kx^2 = x+k

These must be equal where the function is defined for two different intervals

This is at the point x=-6  so let x=-6

k(-6)^2 = -6+k

36k = -6+k

Subtract k from each side

36k-k = -6+k-k

35k = -6

Divide by 35

35k/35 = -6/35

k = -6/35


3 0
3 years ago
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Dvinal [7]

Answer: not sure

but I need points

Step-by-step explanation:

JUST KIDDING It’s A I did this in school before and got it correct :)

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2 years ago
The geometric mean of 4 and 16 is ______. 64 16 12 8
Leni [432]
The geometric mean of x and y is \sqrt{xy}.

So the geometric mean of 4 and 16 is \sqrt{4\times16}=\sqrt{64}=8.
3 0
3 years ago
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