Answer:
a:c = 35:24
a:c = 20:27
a:c = 35:22
a:c = 28:27
Step-by-step explanation:
a:b = 7:3
Using cross products
3a = 7b
Divide by 7
3a/7 = b
Now we want
8b = 5c
Substitute in 3a/7 for b
8 (3a/7) = 5c
24/7a = 5c
Multiply by 7
24/7a *7 = 5*7c
24a = 35c
Divide by c
24 a/c = 35
Divide by 24
a/c = 35/24
a:c = 35:24
a:b = 4:9
Using cross products
9a = 4b
Divide by 4
9a/4 = b
Now we want
3b = 5c
Substitute in 9a/4 for b
3 (9a/4) = 5c
27/4a = 5c
Multiply by 4
27/4a *4 = 5*4c
27a = 20c
Divide by c
27 a/c = 20
Divide by 27
a/c = 20/27
a:c = 20:27
b:c = 5:11
Using cross products
11b = 5c
Divide by 11
b = 5c/11
Now we want
2a = 7b
Substitute in 5c/11 for b
2a = 7(5c/11)
2a = 35c/11
Multiply by 11
2a*11 = 35c
22a = 35c
Divide by c
22 a/c = 35
Divide by 22
a/c = 35/22
a:c = 35:22
b:c = 14:3
Using cross products
3b = 14c
Divide by 3
b = 14c/3
Now we want
9a = 2b
Substitute in 14c/3 for b
9a = 2(14c/3)
9a = 28c/3
Multiply by 3
9a*3 = 28c
27a = 28c
Divide by c
27 a/c = 28
Divide by 27
a/c = 28/27
a:c = 28:27
Answer: (0,-1/2)
Step-by-step explanation:
Edge :-)
Hello,
x²/25+y²/9=1
a=5, b=3 c²=a²-b²=25-9=16=4²
Foci are (0,4) and (0,-4)
Answer C
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Answer:
(5a−3)^2
Step-by-step explanation:
25a^2 - 30a + 9
Factor the expression by grouping. First, the expression needs to be rewritten as 25a^2+pa+qa+9. To find p and q, set up a system to be solved.
p+q=−30
pq=25×9=225
Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 225.
−1,−225
−3,−75
−5,−45
−9,−25
−15,−15
Calculate the sum for each pair.
−1−225=−226
−3−75=−78
−5−45=−50
−9−25=−34
−15−15=−30
The solution is the pair that gives sum −30.
p=−15
q=−15
Rewrite 25a^2 - 30a + 9 as (25a^2−15a)+(−15a+9).
(25a^2−15a)+(−15a+9)
Factor out 5a in the first and −3 in the second group.
5a(5a−3)−3(5a−3)
Factor out common term 5a−3 by using distributive property.
(5a−3)(5a−3)
Rewrite as a binomial square.
(5a−3)^2
