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xenn [34]
3 years ago
12

Solve the system of equations.

Mathematics
2 answers:
Leno4ka [110]3 years ago
7 0

Answer:

x = 7

y =  0

Step-by-step explanation:

Substituting x = - 6y + 7 in 2x - 9y = 14, we get:

= 2(- 6y + 7) - 9y = 14

= - 12y + 14 - 9y = 14

= - 21y = 14 - 14

= y = 0

Putting y = 0 in x = - 6y + 7, we get:

= x = - 6 * (0) + 7

x = 7

Hope this helps!

Alenkasestr [34]3 years ago
4 0

Step-by-step explanation:

<u>Step 1:  Substitute x from the second equation into the second one</u>

2x - 9y = 14

2(-6y + 7) - 9y = 14

(2 * -6y) + (2 * 7) - 9y = 14

-12y + 14 - 9y = 14

-21y +14 - 14 = 14 - 14

-21y/-21 = 0 / -21

y = 0

<u>Step 2:  Substitute y into the second equation</u>

x = -6y + 7

x = -6(0) + 7

x = 0 + 7

x = 7

Answer:  x = 7, y = 0

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jolli1 [7]

Answer:

m>a is 40

Step-by-step explanation:

4x + 50 = 90

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x = 10

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3 0
3 years ago
A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple
tester [92]

Answer:  The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation:  We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.

Therefore, the probability of event A is given by

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

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Answer:

-2

Step-by-step explanation:

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