The fraction 13/25 is equivalent to 52 percent.
To workout this, divide the numerator by the denominator, sum this value to the integer part and multiply the result by 100%, so:
13/25 in percent = 13 ÷ 25 × 100% = 0.52 × 100% = 52%
Answer:
2x^2+4x^2+11x
Step-by-step explanation:
Answer:
50x, x being months.
Step-by-step explanation:
After 2 years, or 24 months, she will have $1,200.
Answer:
Area of this parallelogram: 35.
Step-by-step explanation:
What's the equation for the area A of a parallelogram?
A = b · h,
where
- b is the length of the base, and
- h is the length of the height <em>on that base</em>.
Plot the four vertices on a cartesian plane. Two opposite sides of this parallelogram are parallel to the x-axis. As a result, the length of the base is the same as the difference in x-coordinates between two vertices on that side. As seen on the diagram, the length of this base is b = 2 - (-5) = 7.
The height on either of those two sides will be normal to the x-axis and parallel to the y-axis. The length of those heights will be the same as the difference in y-coordinates between two vertices, one on each line. As seen on the diagram, the length of the height on base <em>b</em> is h = 4 - (-1) = 5.
Area of this parallelogram:
A = b · h = 5 × 7 = 35.
Let P be Brandon's starting point and Q be the point directly across the river from P.
<span>Now let R be the point where Brandon swims to on the opposite shore, and let </span>
<span>QR = x. Then he will swim a distance of sqrt(50^2 + x^2) meters and then run </span>
<span>a distance of (300 - x) meters. Since time = distance/speed, the time of travel T is </span>
<span>T = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x). Now differentiate with respect to x: </span>
<span>dT/dx = (1/4)*(2500 + x^2)^(-1/2) *(2x) - (1/6). Now to find the critical points set </span>
<span>dT/dx = 0, which will be the case when </span>
<span>(x/2) / sqrt(2500 + x^2) = 1/6 ----> </span>
<span>3x = sqrt(2500 + x^2) ----> </span>
<span>9x^2 = 2500 + x^2 ----> 8x^2 = 2500 ---> x^2 = 625/2 ---> x = (25/2)*sqrt(2) m, </span>
<span>which is about 17.7 m downstream from Q. </span>
<span>Now d/dx(dT/dx) = 1250/(2500 + x^2) > 0 for x = 17.7, so by the second derivative </span>
<span>test the time of travel, T, is minimized at x = (25/2)*sqrt(2) m. So to find the </span>
<span>minimum travel time just plug this value of x into to equation for T: </span>
<span>T(x) = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x) ----> </span>
<span>T((25/2)*sqrt(2)) = (1/2)*(sqrt(2500 + (625/2)) + (1/6)*(300 - (25/2)*sqrt(2)) = 73.57 s.</span><span>
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</span><span>mind blown</span>
