How long will it take you to drive 175 miles at a speed of 25 miles per hour

Mathematics

1 answer:

motikmotik2 years ago

5 0

Seven hours because if you think of it as quarters

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Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

3 years ago

Read 2 more answers

Find the value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2

soldier1979 [14.2K]

The value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2 is -2

Solution:

Given that line is passing through point (-5, 2) and (3, r)

Slope of the line is $\frac{-1}{2}$

Need to determine value of r.

Slope of a line passing through point $\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)$ is given by following formula:

$\text { Slope } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ --- eqn 1

$\text { In our case } x_{1}=-5, y_{1}=2, x_{2}=3, y_{2}=\mathrm{r} \text { and } m=-\frac{1}{2}$

On substituting the given value in (1) we get

$\begin{array}{l}{-\frac{1}{2}=\frac{r-2}{3-(-5)}} \\\\ {\text { Solving the above expression to get value of } r} \\\\ {=>-\frac{1}{2}=\frac{r-2}{3+5}} \\\\ {=>-8=\frac{r-2}{3+5}} \\\\ {=>-8=2(r-2)} \\\\ {=>-8=2 r-4} \\\\ {=>2 r=-8+4} \\\\ {=>2 r=-4} \\\\ {=>r=\frac{-4}{2}=-2}\end{array}$