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zmey [24]
3 years ago
14

Please help me. This was my study page but this was hard.

Mathematics
1 answer:
iVinArrow [24]3 years ago
3 0
Your answer is 3277 because a negative divided by a negitive  makes it a positive.

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3<br> Solve for x. Round answer to the nearest tenth.<br> X<br> 759<br> 17
QveST [7]

Answer:

x ≈ 4.6

Step-by-step explanation:

Reference angle = 75°

Opposite side to reference angle = 17

Adjacent side = x

Applying TOA:

Tan 75 = opp/adj

Tan 75 = 17/x

Multiply both sides by x

x*Tan 75 = 17

Divide both sides by Tan 75

x = 17/Tan 75

x ≈ 4.6

7 0
3 years ago
Rachel enjoys exercising outdoors. Today she walked 5 2/3 miles in 2 2/3 hours. What is Rachel’s unit walking rate in miles per
Dmitriy789 [7]
5 2/3 divided by 2 2/3 = 17/3 divided by 8/3 = 17/8 = 2 1/8 miles per hour

2 2/3 divided by 5 2/3 = 8/3 divided by 17/3 = 8/17 hours per mile
5 0
2 years ago
Read 2 more answers
In order to solve the following system of equations by addition, which of the following could you do before adding the equations
alexandr402 [8]
To solve the question, the both equations should have a positive value and a negative value with same number. Then one variable can be eliminated.

To eliminate X term first, we should multiply the first equation by 2 and second one by 3.
Then,
     6X - 12Y = -30
     -6X + 15Y = 70

Or

To eliminate Y term first, we should multiply the first equation by 5 and second one by 6.
 Then,
     15X - 30Y = -75
      -12X + 30Y = 84

According to the choices given 'A' is the correct answer.

5 0
3 years ago
Will give brainliest FAST
sleet_krkn [62]
This is the answer I wish you will understand

7 0
3 years ago
A box at a yard sale contains 3 different china dinner sets, each consisting of 5 plates. A customer will randomly select 2 plat
marshall27 [118]

Answer:

\dfrac{2}{7}

Step-by-step explanation:

3 different china dinner sets, each consisting of 5 plates consist of 15 plates.

A customer can select 2 plates in

C^{15}_2=\dfrac{15!}{2!(15-2)!}=\dfrac{15!}{13!\cdot 2!}=\dfrac{13!\cdot 14\cdot 15}{2\cdot 13!}=7\cdot 15=105

different ways.

2 plates can be selected from the same dinner set in

3\cdot C^5_2=3\cdot \dfrac{5!}{2!(5-2)!}=3\cdot \dfrac{3!\cdot 4\cdot 5}{2\cdot 3!}=3\cdot 2\cdot 5=30

different ways.

Thus, the probability that the 2 plates selected will be from the same dinner set is

Pr=\dfrac{30}{105}=\dfrac{6}{21}=\dfrac{2}{7}

7 0
3 years ago
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