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2 years ago

Q4 Find the missing angle x

Mathematics

1 answer:

BaLLatris [955]2 years ago

5 0

Answer:

Step-by-step explanation:

X = 1/2 arc QS

arc QS is

360 =145+105+QS

110 = QS

so X is 1/2 QS

X= 55°

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In the figure, angle D measures 45° and angle E measures 87°. What is the measurement of angle H? 42°

Vinvika [58]

The complete question is shown in the attached image

Answer:
measure angle H = 42°

Explanation:
There is a theorem that states that:
"In a triangle, the measure of the exterior angle is equal to the sum of the measures of the two non-adjacent angles to the exterior angle"

In the given, we have:
angle E is an exterior angle to triangle DFH
The two non-adjacent angles to angle E are angles D and H

Based on the above rule:
measure angle E = measure angle D + measure angle H

We are given that:
angle E = 87°
angle D = 45°

Substitute with the givens in the above rule and solve for the measure of angle H as follows:
angle E = angle D + angle H
87° = 45° + measure angle H
measure angle H = 87° - 45°
measure angle H = 42°

Hope this helps :)

3 0

3 years ago

I need help, I don't know how to do it.

Sindrei [870]

Answer:

Step-by-step explanation:

to solve this problem we can use the Pythagorean theorem

UT and TL are the legs, while LU is the hypotenuse

We have to find LU so we can proceed like this

x^2 + (x+1)^2 = LU^2

x^2 + x^2 + 1 + 2x = LU^2

2x^2 + 2x + 1 = LU^2

LU = +/- $\sqrt{2x^2+2x+1$

we have to take only the positive value because a length can’t be negative.

2x^2 + 2x + 1 is positive for every value of x, so the final answer is

$\sqrt{2x^2+2x+1}$

5 0

2 years ago

Plssssss help fast!!!!Launching From a Tower

aleksandrvk [35]

Answer:

A. h = 64

B. t = 4 -> 4 seconds

C. t = 1.5 -> 1.5 seconds

D. maximum height is 100 ft

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

Step-by-step explanation:

h = -16t2 + vt + 64

A. What tower platform height was the projectile launched from?

when the projectile was not launched, t = 0

h = -16(0)^2 + v(0) + 64 = 64

B. How long was the projectile in the air?

if the projectile lands, its height = 0 so substitute 0 for h

0 = -16(t)^2 + 48(t) + 64

= -16(t^2 - 3t - 4)

= -16(t - 4)(t + 1)

t = 4 or t = -1

Since time can not be negative, t = -1 can not be the answer. Therefore, the projectile lands when t = 4 or 4 seconds.

C. When did it reach its maximum height?

maximum -> t=-b/2a where in -16(t)^2 + 48(t) + 64, b = 48 and a = -16

t = -48/-32 = 1.5

D. What was its maximum height?

plug t = 1.5 into -16(t)^2 + 48(t) + 64

-16(1.5)^2 + 48(1.5) + 64 = -36 + 72 + 64 = 100

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

6 0

2 years ago

Give the following equations determine if the lines are parallel perpendicular or neither

GaryK [48]

In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.

Let's start with the first equation.