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zhuklara [117]
3 years ago
10

Melissa the trainer has two solo workout plans that she offers her clients: Plan A and Plan B. Each client does either one or th

e other (not both). On Wednesday there were 5 clients who did Plan A and 3 who did Plan B. On Thursday there were 2 clients who did Plan A and 12 who did Plan B. Melissa trained her Wednesday clients for a total of 7 hours and her Thursday clients for a total of 19 hours. How long does each of the workout plans last? LengthofeachPlanAworkout:hour(s) LengthofeachPlanBworkout:hour(s)
Mathematics
1 answer:
Vikentia [17]3 years ago
6 0

Let Length of each Plan A workout be a

and Length of each Plan B workout be $b$

on Wednesday,

$5a+3b=7$

$2a+12b=19$

multiply equation one by $4$ and subtract equation two from it

to get,

$18a=9$ or $a=\frac12$

substitute $a$ in eq 2. to get $b$, $12b=18\implies b=\frac32$

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6 0
3 years ago
3x+5=19-4x<br>What does x equal?​
alexandr1967 [171]

Answer:

x=2

Step-by-step explanation:

x=2 in the equation 3x+5=19-4x

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3 years ago
Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention
Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

a=\frac{d^2s}{dt^2}

Differentiate the position vector with respect to t.

\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}

Differentiate both sides of the obtained equation with respect to t.

\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}

The initial position is obtained at t=0. Substitute t=0 in the given position function.

\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}

8 0
1 year ago
In Triangle ABC shown below, side AB is 6 and side AC is 4.
Amanda [17]

Answer: A is the correct option.Segment AD is 3 and segment AE is 2.


Step-by-step explanation:

Given : A triangle ABC  where AC=4 and AB=6

then to prove segment DE is parallel to segment BC and half its length.

the length of AD and AE must divide AC and AB respectively to get the same ratio of 2:1

To apply converse of basic proportionality theorem.

If we take first option Segment AD is 3 and segment AE is 2 then

\frac{AB}{AD}=\frac{6}{3}=\frac{2}{1},\frac{AC}{AE}=\frac{4}{2}=\frac{2}{1}\\\Rightarrow\frac{AB}{AD}=\frac{AC}{AE}

Therefore by converse of basic proportionality theorem

DE is parallel to segment BC and half its length.

Therefore A is correct option.




7 0
3 years ago
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Fourteen thousand seven hundred ninety three in standard form
NeTakaya

Answer:

14,793

Step-by-step explanation:

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3 years ago
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