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Ahat [919]
2 years ago
13

Raphael paid $158 for a camera during a 75% off sale. what was the cameras regular price?

Mathematics
2 answers:
skad [1K]2 years ago
8 0
Regular price - (regular price × discount) = sale price
$632 - ($632 × 75%)
$632 - $474
$158
regular price = $632
Serga [27]2 years ago
6 0
$210.67 is the answer. Rounded, of course, from 210.666666667
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Alex wrote the expanded form for the number 165.038 as “100 + 60 + 5 + 30 + 8.” Was he correct? If not, give the correct expande
ivann1987 [24]

No, Alex is wrong. The correct expanded form would be 100+60+5+.03+.008

8 0
2 years ago
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Are the coordinates (-3,7),(0,4),(2,0), and (6,-4) on the same line?
Advocard [28]

Answer:

No

Step-by-step explanation:

7 0
2 years ago
Sketch the graph of 4x² + 25x – 21 = y using your graphing calculator. What are the x-intercepts of this graph
denis-greek [22]

Answer:

B .75 - 7

Step-by-step explanation:

3 0
3 years ago
The lines shown below are perpendicular. if the green line has a slope of 3/4, what is the slope of the red line
devlian [24]
Perpendicular slope is opposite and reciprocal

<span>the green line has a slope of 3/4
so </span><span>the slope of the red line is -4/3

answer 
</span><span>c)-4/3</span>
6 0
3 years ago
Read 2 more answers
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex
azamat

Answer:

Maximum at points (8,0),(-8,0).Minimum at points (0,8), (0,-8).

Step-by-step explanation:

There are multiple ways of using lagrange multipliers. Most of them are equivalent.

Consider the function F(x,y) = x^2-y^2-\lambda(x^2+y^2-64). We want the following \frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial \lambda} = 0.

Then, we have

\frac{\partial F}{\partial x} = 2x-2x\lambda= 2x(1-\lambda)=0

\frac{\partial F}{\partial y} = -2y-2y\lambda = -2y(1+\lambda)=0

\frac{\partial F}{\partial \lambda} = x^2+y^2-64=0

From the first two equations, we can see that if \lambda =1 then necessarily y=0. IN that case, from the third equation (which is the restriction) gives us that x=\pm 8.

On the other hand, if \lambda=-1 then necessarily x=0. Again, using the restriction this gives us that y=\pm 8.

if we evaluate the original function in this points, we have that f(0,\pm 8) = -64, f(\pm 8,0)=64. Then, we have Maximum at points (8,0),(-8,0) and Minimum at points (0,8), (0,-8).

3 0
3 years ago
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