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3 years ago

How could i find the missing lengths?

Mathematics

1 answer:

zysi [14]3 years ago

5 0

You could use the Pythagorean Theorem
${a}^{2} + {b}^{2} = {c}^{2}$

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What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?

kompoz [17]

Answer:

K(-4,8) is the ortho center.

Step-by-step explanation:

In a right angled triangle, The vertex of the right angle is the ortho center.

Here we are given

J(-4,-1), K(-4,8) & L(2,8)

Using distance formula we get

$JK^{2}= (-4+4)^{2}+ (8+1)^{2}=0+81=81$

$KL^{2}= (-4-2)^{2}+ (8+-8)^{2}=36+0=36$

$JL^{2}= (-4-2)^{2}+ (8+1)^{2}=36+81=117$

So we can say that

$JK^2 + KL^2=JL^2$

By converse of pythagorean theorem we get

Hence the Vertex of the right angle is K(-4,8)

K(-4,8) is the ortho center.

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3 years ago

Plz help! Answer at least one of these questions and i will give you brainly! It’s ok if you don’t know them all!

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3 years ago

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Find the equation of the line using the point-slope formula. Write the final equation using slope-intercept form. (1,2) with a s

Lilit [14]

Answer:

$y-2=\displaystyle -\frac{3}{4}(x-1)$

$y=\displaystyle -\frac{3}{4}x+\frac{11}{4}$

Step-by-step explanation:

Hi there!

Point-slope form: $y-y_1=m(x-x_1)$ where m is the slope of the line and $(x_1,y_1)$ is a given point

Given that the slope is -3/4, we can plug it into $y-y_1=m(x-x_1)$ as m:

$y-y_1=\displaystyle -\frac{3}{4}(x-x_1)$

We can also plug in the given point (1,2):

$y-2=\displaystyle -\frac{3}{4}(x-1)$

Slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)

To write the equation in slope-intercept form, isolate y:

$y-2=\displaystyle -\frac{3}{4}(x-1)\\\\y=\displaystyle -\frac{3}{4}(x-1)+2\\\\y=\displaystyle -\frac{3}{4}x+\frac{3}{4}+2\\\\y=\displaystyle -\frac{3}{4}x+\frac{11}{4}$

I hope this helps!

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3 years ago

P=3h+3u solve for h ...?

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h = p/3 - u

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