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2 years ago

For the following system, use the second equation to make a substitution for x in the first equation.

Mathematics

2 answers:

Grace [21]2 years ago

6 0

Answer:

Step-by-step explanation:

slamgirl [31]2 years ago

3 0

A) x + 5y - 10 = 0
B) x = 2y - 8
Substituting B into A
A) 2y -8 + 5y - 10 = 0
A) 7y -18 = 0

7y - 18 = 0
answer is C

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Help me please my test is tomorrow!

Marysya12 [62]

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$4z + 2 = 10z - 1$

Add sides 1

$4z + 2 + 1 = 10z - 1 + 1$

$4z + 3 = 10z$

Subtract sides 4z

$- 4z + 4z + 3 = - 4z + 10z$

$3 = 6z$

$6z = 3$

Divide sides by 6

$\frac{6z}{6} = \frac{3}{6} \\$

$z = \frac{1}{2} \\$

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$3y = y + 3$

Subtract sides y

$- y + 3y = - y + y + 3$

$2y = 3$

Divide sides by 2

$\frac{2y}{2} = \frac{3}{2} \\$

$y = \frac{3}{2} \\$

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Thus the correct answer is :

$a) \: y = \frac{3}{2} \: , \: z = \frac{1}{2} \\$

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3 years ago

To convert a repeating decimal to a fraction, your goal is to:

Nezavi [6.7K]

To convert a repeating decimal to a fraction, your goal is to get the numbers to the left of the decimal point to be the same

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2 years ago

What number should be added to both sides of the following equation to solve for x?

son4ous [18]

Answer:

7 is your answer

Step-by-step explanation:

Hope that helps you

5 0

3 years ago

Read 2 more answers

Muad can lift 129.6 pounds and Samara can lift 135.25 pounds. How much more weight can Samara lift?

notka56 [123]

Answer:5.29 pounds

Step-by-step explanation:

135.25 -129.6

6 0

2 years ago

Read 2 more answers

What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?

olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

$\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}$

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒ 1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒ 1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0

2 years ago