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noname [10]
3 years ago
14

I need help on my math!!!!!!!

Mathematics
1 answer:
AURORKA [14]3 years ago
4 0

Answer:

67

Step-by-step explanation:

GOOD LUCK!!

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PLEASE I need help on these 4 problems you don’t have to do them all but if you can at least do one of them TYSM
Brrunno [24]

Answer:

For right angle triangle,

we use Pythagoras theorem that is:

c^{2} =a^{2} +b^{2}

c = \sqrt{a^{2} +b^{2} }

For question 1:

c = ?

a = 40

b = 9

putting them in formula,

c = \sqrt{40^{2} + 9^{2} }

c = 41

For question 2:

c = ?

a = 12

b = 13

putting them in formula,

c = \sqrt{12^{2} + 13^{2} }

c = approximately 17.69181

For question 3:

c = 35

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

35^{2} = 20^{2} + b^{2}

1225 = 400 + b^{2}

b^{2} = 1225 - 400

b^{2} = 825

\sqrt{b^{2} } = \sqrt{825}

b = 5 \sqrt{33}

For question 4:

c = 37

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

37^{2} = 20^{2} + b^{2}

1369 = 400 + b^{2}

b^{2} = 1369 - 400

b^{2} = 969

Taking square root on both sides

b = 31.12

Hope it helps.

4 0
2 years ago
Christian wants to make an A in his class. There are 5 exams this year, and he has taken 4 of them. His grades were 84, 88, 95,
ioda

Answer:

last test grade=x     x ≥ 79 so 79%

Step-by-step explanation:

(84/5)+(88/5)+(95/5)+(104/5)+(x/5) ≥ 90

Simplify: (371/5)+(x/5) ≥ 90

Subtract (371/5) from both sides

Then you have (x/5) ≥ (79/5)

Then multiply both sides by 5 and you have x ≥ 79

P.S. If it was helpful please consider giving me brainliest. =)

3 0
3 years ago
Explane why kequally 10 is the consentof proporitomally
Oxana [17]
I don't understand the question, could you please re-word it and if I know it I will give u the answer
5 0
3 years ago
Whats 1 and 2<br> true<br> false<br> true/false
slavikrds [6]

Answer:false true

Step-by-step explanation:

4 0
2 years ago
Need help finding the x, y, and z. please and thank you
Reptile [31]

The first step in any problem is to look at what you are given. When solving systems of linear equations, it is often helpful to eliminate one or more of the variables by adding or subtracting a multiple of one equation with a multiple of another. It is convenient when at least one of the multipliers is 1.

Here, we can cancel the y-terms in the 2nd and 3rd equations simply by adding them together. This gives

... (5x +y -4z) +(-3x -y +5z) = (41) +(-45)

... 2x +z = -4 . . . . simplified

Likewise, we can add 3 times the second equation to the first to cancel y in that sum.

... 3(5x +y -4z) +(2x -3y +z) = 3(41) + (-1)

...15x +3y -12z +2x -3y +z = 123 -1

...17x -11z = 122 . . . . simplified

Now that we have 2 equations in x and z, we can go through the same process. We observe that the coefficient of z is +1 in the first equation and -11 in the second. The means we can cancel the z terms by adding 11 times the first equation to the second:

... 11(2x +z) +(17x -11z) = 11(-4) +122

...22x +17x = 78 . . . . . . simplify a little bit

... x = 78/39 = 2 . . . . . . divide by 39

From above, we find

... z = -4 -2x = -4 -2·2 = -8

... y = 41 -5x +4z = 41-5(2) +4(-8) = 41 -42 = -1

The solution is (x, y, z) = (2, -1, -8).

_____

The method of elimination used here will vary with the system of equations. If you want to employ a consistent method, you can use Cramer's Rule, Gaussian elimination, or matrix methods. Since you apparently don't mind help from technology, learning to do this on your graphing calculator can also be a good idea.

4 0
3 years ago
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