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3 years ago

HELP ME PLEASE! In triangle FGH, GH=1, FG=12, and GH=sq. root of 145

Mathematics

1 answer:

topjm [15]3 years ago

3 0

Answer:

B&C: tan F=0.0833 & tan H=12.0000

Step-by-step explanation:

The tangent relationship is always opposite/adjacent.

When F is the reference angle, side GH (1) is the opposite side and side FG (12) is the adjacent.

1/12 is 0.0833

When H is the reference angle, FG (12) is the opposite and GH (1) is the adjacent.

12/1 is 12.

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Vinvika [58]

Answer:

bx=-3

Step-by-step explanation:

-2(bx-5)=16

Use distributive property.

-2bx+10=16

Now, subtract 10 from both sides.

-2bx=6

Divide -2 from both sides.

bx=-3

Hope this helps!

If not, I am sorry.

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2 years ago

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Peter is 2 years older than Winnie. Peter's age is 16 years less than seven times Winnie's age. The equations below model the re

Nataliya [291]

I would say P = w + 2, because their ages change over time but the difference between the ages will always be the same.

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3 years ago

NO LINKS!!! Please help me with these problems

nadezda [96]

<h3>Answers:</h3>

7) Center= (-1,2) Radius= $\boldsymbol{\sqrt{8}}$ Equation: $(x+1)^2+(y-2)^2 = 8$

8) Center= (3,13) Radius= 13 Equation: $(x-3)^2+(y-13)^2 = 169$

=========================================================

Explanation:

Problem 7

Let's find the distance from (-1,2) to (-3,4)

$(x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\$

This is the radius because it stretches from the center to a point on the circle, so $r = \sqrt{8}$

Squaring both sides will get us $r^2 = 8$

One useful template for a circle is the equation $(x-h)^2+(y-k)^2 = r^2\\\\$

(h,k) is the center

r is the radius

Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.

$(x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\$

You can confirm this by using a tool like Desmos. See below.

------------------------------------------------------------------------

Problem 8

The endpoints of the diameter are (-2,1) and (8,25)

The center is the midpoint of these endpoints.

The midpoint of the x coordinates is (-2+8)/2 = 3

The midpoint of the y coordinates is (1+25)/2 = 13

The center is (h,k) = (3,13)

Now find the distance from the center to one of the points on the circle, let's say to (8,25)

$(x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\$