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Korvikt [17]
3 years ago
11

Below are two parallel lines with a third line intersecting them.

Mathematics
1 answer:
lesya [120]3 years ago
6 0

Answer:

x = 77

Step-by-step explanation:

no hablo ingles lo siento

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116.PS-9<br> Question Help<br> Find the surface area of the prism.<br> 8 cm<br> 5 cm13cm<br> 12 cm
Ivanshal [37]

Answer:

6,240 cm

Step-by-step explanation:

8 cm x 5 cm x 13 cm x 12 cm = 6,240 cm

8 0
2 years ago
14. Is x + 2 a factor of the polynomial f(x)=2x-3x²-4x+1?
defon
Thanks for the free 10x+ points .
8 0
2 years ago
A quadrilateral has vertices A(4, 5), B(2, 4), Q4, 3), and D(6, 4). Which statement about the quadrilateral is true?
Igoryamba

The true statement about the quadrilateral is ABCD is a rhombus with non-perpendicular adjacent sides.

<h3>How to determine the true statement?</h3>

The coordinates are given as:

A(4, 5), B(2, 4), Q4, 3), and D(6, 4).

Next, we plot the graph of the quadrilateral ABCD

See attachment

From the attached graph, we have the following highlights:

  • Opposite sides are equal
  • Opposite sides are parallel
  • Adjacent sides are not perpendicular

This means that the quadrilateral is ABCD is a rhombus

Read more about quadrilateral at

brainly.com/question/4812739

#SPJ1

<u>Complete question</u>

A quadrilateral has vertices A(4, 5), B(2, 4), C(4, 3), and D(6, 4). Which statement about the quadrilateral is true?

1) ABCD is a parallelogram with noncongruent adjacent sides.

2) ABCD is a trapezoid with only one pair of parallel sides.

3) ABCD is a rectangle with noncongruent adjacent sides.

4) ABCD is a square.

5) ABCD is a rhombus with non-perpendicular adjacent sides.

7 0
1 year ago
Which equation could be used to
Vika [28.1K]
D sounds like the best answer math can be hard tho
3 0
3 years ago
Brandon is on one side of a river that is 50 m wide and wants to reach a point 300 m downstream on the opposite side as quickly
AlexFokin [52]
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>
</span><span>
</span><span>
</span><span>
</span><span>mind blown</span>
8 0
3 years ago
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