Answer:

Step-by-step explanation:
Given

Required
Evaluate

Start by dividing 15 by 3

Apply the following law of indices on indices x and z

So, the expression becomes:



If he wants to earn $1000, it means that $1000 must be 20% of his overall sales.
1000 = 20% * x
1000 = 20/100x
1000 = 0,2x / ÷ 0,2 (both sides)
x = 5000
Answer: A salesman must sell products for $5000 to earn $1000.
<span>Let x be the
width of the frame.
Area of the picture is 5*7 = 35 sq in</span>
Area of the
frame given as the same, frame = 35 sq in also
<span>
Therefore the total area (picture & frame) will be 70 sq in</span>
<span>
A simple equation would be made and it will look like this: (7+2x) * (5+2x) =
70
</span>
<span>Using the FOIL
method
35 + 14x + 10x + 4x^2 = 70
4x^2 + 24x + 35 - 70 = 0; subtract 70 from both sides:
4x^2 + 24x - 35 = 0</span>
<span>
unluckily this will not factor easily, so we should use the quadratic equation:
where: a = 4; b = 24; c = -35</span><span>
</span><span> x = -b sqrt b^2 – 4*a*c divided by 2 * a</span><span>
x
= -24</span> ±<span>
sqrt 24^2 -4*4*(-35) divided by 2*4
x = -24 </span>±<span>sqrt 576 -
-(560) divided by 8
x = -24 </span>± sqrt 576 + 560 divided by 8<span>
x = -24 </span>± sqrt1136
divided by 8<span>
x = -24 + 33.7046 divided by 8
x = 9.7046 divided by 8
x =1.213 inches is the answer</span>
<span> </span>
12x6.5x1.25=97.5 cubic centimeters answer B
Answer:
0.04,0.25.0.52
Step-by-step explanation:
Given that you throw a dart at a circular target of radius 10 inches.
Assuming that you hit the target and that the coordinates of the outcomes are chosen at random,
probability that the dart falls
(a) within 2 inches of the center
Here favourable region has area of a circle with radius 2 inches and sample space has area of 10 inches
Prob = 
(b) within 2 inches of the rim.
For within two inches from the rim we have to select area of the ring i.e. area of big circle with 10 inches - area of smaller circle with 10-2 inches
Prob= 
c) within I quadrant
area of I quadrant / area of circle=0.25
d) within I quadrant and within 2 inches of the rim
= I quadrant area + 2 inches ring area - common area
= 