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Ksju [112]
4 years ago
8

what is the answer to this problem. lily built an orange shelf and a brown shelf to hang in her room. the orange shelf is 5 feet

and 6 inches long. the brown shelf is twice as long. What is the lengh of the btown shelf
Mathematics
1 answer:
marishachu [46]4 years ago
4 0
The orange shelf was 5'6. First, let's put that into inches. There are 12 inches in one foot, and 5 feet. 12x5= 60. 60+6=66 inches. We then Multiply that by 2. 66x2= 132. The brown shelf is 132 inches long. We then put that back into feet. 132/12= 11. The brown shelf is 11 feet long.
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You have a babysitting job that pays $6 an hour. You also have
Natalka [10]

Answer:

One possibility is to work for (10) hours as a babysitter, and (10) hours as a cashier.

Step-by-step explanation:

An easy way to solve this problem is to set up a system to model the situation. Create one equation to model the money make, and the other to model the time spent. Let parameters (x) and (y) represent the time one spends at each job.

Since one cannot spend more than (20) hours a week working, set the first equation, for time, equal to (20),

x + y = 20

Now multiply each unit for the time by the money earned at each job, set this new equation equal to (150), the minimum amount of money one wishes to earn,

6(x) + 9(y) = 150

Thus the system is the following,

\left \{ {{x+y=20} \atop {6x+9y=150}} \right.

Now use the process of elimination. The process of elimination is when one multiplies one of the equations by a term such that when one adds the two equations, one of the variables cancels. One can solve for the other variable, and then backsolve for the first variable. Multiply the first equation by (-6) so that the variable (x) cancels.

\left \{ {{-6x-6y=-120} \atop {6x+9y=150}} \right.

Add the two equations,

3y=30

Use inverse operations to solve for (y),

y=10

Now substitute the value of (y) back into one of the original equations and solve for (x),

x+y=20

x+10=20

x=10

6 0
3 years ago
The given line segment has a midpoint at (3, 1). What is the equation, in slope-intercept form, of the perpendicular bisector of
viva [34]

Answer:

Step-by-step explanation:

I think the attached photo supports for your question

Here is my anser:

We need to find the slope of the of and from the graph, we see that if x increases from 2 to 4, y decreases from 4 to -2.  Thus, the slope of the blue line is : \frac{-2-4}{4-2} = -3

But the slope of the perpend. bisector of the blue line is the negative reciprocal of -3, or  m= 1/3.

Let's find the slope-intercept form of this bisector.  We need to determine b in y=mx+b.  Referring to the midpoint of the blue line, x= 3; y= 1; and m=1/3.  Then

y=mx+b becomes 1=(1/3)(3) + b.  Solving for b:  1=1+b.  Then b=0.

Thus, the equation of the perpendicular bisector of the blue line through (3,1) is  y=mx + b, or y=(1/3)x + 0, or y=x/3.

9 0
3 years ago
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If 8n = -40, then n =​
julia-pushkina [17]

Answer:

n=-5

Step-by-step explanation:

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4.6 2.8 17 7 least to greatest ​
o-na [289]

Answer:

2.8, 4.6, 7, 17

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The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of th
harkovskaia [24]

Answer:

The displacement of the object on this intervals is 1.33 m.

Step-by-step explanation:

Given that,

The function of velocity is

v=\dfrac{1}{2t+4}\ m/s

For 0 ≤ t ≤8 , n = 2

We need to calculate the intervals

Using formula for intervals

For, n = 1

\Delta x=\dfrac{t_{f}-t_{i}}{n}

\Delta x=\dfrac{8-0}{2}

\Delta x=4

So, The intervals are (0,4), (4,8)

We need to calculate the velocity

Using given function

v=\dfrac{1}{2t+4}

For first interval (0,4),

Put the value into the formula

v_{0}=\dfrac{1}{2\times0+4}

v_{0}=\dfrac{1}{4}

For first interval (4,8),

Put the value into the formula

v_{4}=\dfrac{1}{2\times4+4}

v_{4}=\dfrac{1}{12}

We need to calculate the total displacement

Using formula of displacement

D=(v_{0}+v_{4})\times(\Delta x)

Put the value into the formula

D=(\dfrac{1}{4}+\dfrac{1}{12})\times4

D=1.33\ m

Hence, The displacement of the object on this intervals is 1.33 m.

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