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Oxana [17]
2 years ago
8

What is the midpoint of the segment shown below?

Mathematics
1 answer:
sashaice [31]2 years ago
3 0

Answer:

Step-by-step explanation:

average of x-coordinates = (-11+9)/2 = -1

average of y-coordinates = (0-1)/2 = -½

midpoint: (-1,-½)

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Guys, help me out, please.
goldfiish [28.3K]

Answer:

4

Step-by-step explanation:

Given

\frac{28}{3+4} ← evaluate the denominator

= \frac{28}{7} ← perform the division

= 4

7 0
3 years ago
Answer all of the questions below in order to determine the volume of the rectangular prism. 10 3 5 Press the play button: play
soldier1979 [14.2K]

Answer:

10, 50, 150

Step-by-step explanation:

i just did it lol.

7 0
3 years ago
Can you help me to find the values of abc and cde ? ​
Agata [3.3K]

Answer:

88° and 132°

Step-by-step explanation:

The sum of angles in a pentagon ( a 5-sided shape) is given as

= (5 - 2) 180°

= 540°

The angles ∠EAB and ∠AED are supplementary hence the sum is 180° Therefore,

∠AED + 110 = 180

∠AED = 180 - 110

= 70°

Given that the sum of the angles in a pentagon is 540° then

110 + 70 + 2k + 140 + 3k = 540

5k + 320 = 540

5k = 540 - 320

5k = 220

k = 220/5

= 44°

Hence the angle ∠ABC

= 2 × 44

= 88°

∠CDE

= 3 × 44

= 132°

4 0
3 years ago
Find the missing length indicated ​
tiny-mole [99]

Answer:

x = 256

Step-by-step explanation:

Remark

Interesting question. Thanks for posting.

The small triangle has the height as the missing side.

Part One: Givens

a = 144

b = height = ?

c = 240

a^2 + b^2 = c^2

144^2 + b^2 = 240^2

20736 + b^2 = 57600                   Subtract 20736 from both sides

b^2 = 57600 - 20736

b^2 = 36864

√b^2 = √36864

b = 192

Solve for x

The relationship between the height (b) and 144 and x is

b^2 = 144 * x

36864 = 144x          Divide by 144

x = 256

5 0
3 years ago
Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
allsm [11]

Answer:

L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]

Step-by-step explanation:

Given that:

f(t)  = 12 cos (t- \dfrac{\pi}{6})

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

f(t) = 12 [cos\  t \  cos \dfrac{\pi}{6}+ sin \ t  \ sin \dfrac{\pi}{6}]

f(t) = 12 [cos \  t \ \dfrac{3}{2}+ sin  \ t  \ sin \dfrac{1}{2}]

f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)

L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]

L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]

L(f(t)) = 6 \sqrt{3}  \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}

L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}

L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}

L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]

7 0
3 years ago
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