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olasank [31]
3 years ago
6

I don’t have a question anymore kskskakakalalalakallakakaka

Mathematics
2 answers:
wlad13 [49]3 years ago
6 0
duh hjdmxbzuksbxjckdvHoxjcjsk i hxjsvgxosbsgxbdkksusbdc
kramer3 years ago
5 0
Ojsjsijdjsndjfjfiowjsbdidhejsnfj
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Suppose c and d vary inversely, and d = 2 when c = 17.
stepladder [879]

Answer:

if d=2 and c=17

then c=68 and d=8

please mark me as brainlist ....

hope e will help you

4 0
3 years ago
hi:) for 5(b) , I got 157.7 & 22.3 but in the answer key, the only correct answer is 157.7. I thought there should be 2 poss
n200080 [17]

Answer:

Θ = 157.7°

Step-by-step explanation:

Given

cosΘ = - 0.925

Since cosΘ < 0 then Θ is in the second or third quadrant.

Since 0° ≤ Θ ≤ 180° then Θ is in the second quadrant. thus

Θ = cos^{-1}(- 0.925) = 157.7° ← angle in second quadrant

8 0
3 years ago
Read 2 more answers
What do you do to find m?
lesya692 [45]
Ok here is how to solve these I hope this helps
7       =      2-5m
+5m           +5m
7+5m =     2
-7               -7
5m      =    -5
/5              /5
m=-1


<span>-Hope this helps :)</span>
8 0
3 years ago
1. If the product of the integers a,b, and c is 1, then what is the difference between the largest and the smallest possible val
Natali5045456 [20]
For question 1, the largest value would be 1 and the smallest value would be -1. A, B, and C are either all 1 or two is negative and one is positive, either way the number is 1 and it's opposite. 1^2 is 1, 1^3 is 1, 1^4 is 1. The product is 1. -1^2 is -1, -1^3 is -1, -1^4 is -1. The product is -1. Now the difference. 1 - (-1), which is 1 + 1, the answer is [D) 2]
4 0
3 years ago
The mean of a set of data is 148.87 and its standard deviation is 68.29. Find the z score for a value of 490.19.
Ksivusya [100]
<span>The mean of a set of data is 148.87 and its standard deviation is 68.29. Find the z score for a value of 490.19
the z-score is given by:
z=(x-</span>μ<span>)/</span>σ
plugging in the values in the expression we get:
z=(490.19-149.87)/68.29
z=340.32/68.29
z=4.9835
6 0
3 years ago
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