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zubka84 [21]
2 years ago
7

Solution to 3x+5y=-20

Mathematics
2 answers:
Wewaii [24]2 years ago
7 0
Answer: 20


3x+5y=-20

5y=-20divide

answer is x=-4

3x=-20 divide

the answer is y=-20



I hope this helps and have a wonderful day filled with joy!!

zloy xaker [14]2 years ago
3 0

3x+5y=-20

5y=-20divide

the answer is x=-4

3x=-20 divide

the answer is y=-20

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Answer:

m=2

Step-by-step explanation:

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2 years ago
The student council has 10 members where 5 of the members are Seniors. They need to choose a President, Vice President, Secretar
natima [27]

Answer:

P=1/42.

Step-by-step explanation:

We know that the student council has 10 members where 5 of the members are Seniors. They need to choose a President, Vice President, Secretary and Treasurer. We calculate the probability that the President is a Senior:

We calculate the number of possible combinations:

C_4^{10}=\frac{10!}{4!(10-4)!}=210

Number of favorable combinations is 5.

Threfore, the probability is

P=5/210

P=1/42.

8 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

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3 years ago
Find the roots of each equation by factoring
Korolek [52]

Answer:

x=7 x=-7

Step-by-step explanation:

Solution is attached.

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