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alexandr402 [8]
3 years ago
14

Need help ON THESE TOO PLEASE WILL MARK BRAINLY

Mathematics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

1)

angle 2WZX = angle ZYX ( Since all angles and sides of a rhombus are equal)

ZYX = 2 x 29.5

59

2)

4x - 10 =90 (since all angles of a rhombus intersect at 90°)

x=90+10/4

x=100/4

x=25

hope this helps

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Solve the system of equations and choose the correct answer from the list of options.
Igoryamba

Answer:

x = -6/5

y = -8/5

Step-by-step explanation:

2x + y = -4

Y = 3x + 2

Substitute the second equation into the first equation

2x + 3x+2 = -4

Combine like terms

5x +2 = -4

Subtract 2 from each side

5x+2-2=-4-2

5x = -6

Divide by 5

5x/5 = -6/5

x = -6/5

Now we need to find y

y = 3x+2

y = 3(-6/5)+2

y = -18/5 + 10/5

y = -8/5

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3 years ago
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Fantom [35]
It’s 16 because you need to kiss the moon
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Lacy runs 5 1/4 km on Thursday. She runs 1 1/2 times as far on Saturday. How far does Lacy run on Saturday? Express your answer
natita [175]

Answer: 5 1/4 x 1 1/5 = 7 1/2 + 3/8 -> 7 7/8

The answer is 7 and 7/8

3 0
3 years ago
HELP PLEASE ASAP
kondaur [170]
Answer: D. $3,450
Given: 2,500 shares of a stock bought for $8 a share
<span>Probabilities:  the stock will rise to a value of $32 a share is 22%</span>  <span>                                                the stock will fall to $3 a share is 78%</span>
<span>Question: What is the expected value of the investor’s profit from buying the stock?</span>
Solution: (22%(32-8)-78%(8-3))2500
                (.22(24)-.78(5))2500
                (5.28-3.9)2500
                (1.38)2500
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4 years ago
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla
forsale [732]

Answer:

a)

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = 3.37*10^{-2} meters per second squared

b)

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = 2.92*10^{-2} meters per second squared

Step-by-step explanation:

The formula is:

v=\frac{2 \pi R}{T}

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

a_c=\frac{4 \pi^{2} R}{T^2}

Now,

a)

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}

Centripetal Acceleration at Equator = 3.37*10^{-2} meters per second squared

b)

At 30.0° north of the equator:

R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6

Now,

Speed = v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}

Centripetal Acceleration at 30 degrees north of equator = 2.92*10^{-2} meters per second squared

4 0
3 years ago
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