All categories

2 years ago

In the diagram, GH bisects FGI. A. Solve for x and find m FGH. B. Find m HGI. C.find m FGI. A. X=_(simplify your answer)

Mathematics

1 answer:

Sophie [7]2 years ago

8 0

Answer:

Part A) x=9 and m∠FGH=29°

Part B) m∠HGI=29°

Part C) m∠FGI=58°

Step-by-step explanation:

Part A) Solve for x and find m∠FGH

we know that

m∠FGI=m∠FGH+m∠HGI

If GH bisects FGI

then

m∠FGH=m∠HGI

substitute the values

4x-7=5x-16

solve for x

5x-4x=16-7

x=9

m∠FGH=4x-7

m∠FGH=4(9)-7=29°

Part B) Find the measure of angle HGI

m∠HGI=5x-16

Remember that

m∠FGH=m∠HGI

Verify

m∠HGI=5(9)-16=29° -----> is correct

Part C) Find the measure of angle FGI

m∠FGI=m∠FGH+m∠HGI

substitute the values

m∠FGI=29°+29°=58°

You might be interested in

How to answer a^3-216

marshall27 [118]

Are you asking to factor it? if so, the answer is (x-6)(x^2+6x+36)

8 0

3 years ago

Please help me out with number 7

Slav-nsk [51]

The answer for number 7 is x=3

4 0

3 years ago

Interpret the slope based on the chart.

bogdanovich [222]

Answer:

b. For 100 additional burglaries in California one expects to see an increase of about 3 burglaries in Hawaii.

Step-by-step explanation:

The slope gives the increase in the independent variable per unit increase in the dependent variable. Hence, a reasonable interpretation for the slope would be ;

the rate of change in the number of burglaries in Hawaii per 100 unit change in number of burglaries in California is 3.

Hence, for 100 additional burglaries in California, the rate of change in the number of burglaries in Hawaii is 3.

3 0

2 years ago

M% of n is what percent of p?

babunello [35]

Let x be the percent of p asked
then
m% of n = x% of p
m/100*n = x/100*p
mn = xp
x = mn/p

8 0

3 years ago

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$

4 0

3 years ago