3 years ago

Solve equations 5-5x=14-8x 3(2y-5)=4y-7

Mathematics

1 answer:

GaryK [48]3 years ago

5 0

5-5x=14-8x
-5x=9-8x
3x=9
x=3

3(2y-5)=4y-7
6y-15=4y-7
2y-15=-7
2y=8
y=4

You might be interested in

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B%20%5Csqrt%5B40404504x%20%5Cfrac%7Bx05121y%20%5Csqrt%5By%20%7C4%7C%20%5D%7B%3F%7D

atroni [7]

Answer:

The answer is 12 |17|7|5

what is the question?

Step-by-step explanation:

stan bts

6 0

2 years ago

Then each week thereafter, you add $5 to the account but no interest is earned. Suppose you receive $100 for a graduation presen

irga5000 [103]

100 + 5x = y

100 +5x = 310
x = 42 weeks

7 0

3 years ago

Read 2 more answers

Find the values of k so that each remainder is three. 10. (x^2+ 5x + 7) = (x + k)

Goryan [66]

Answer:

$k=1\text{ or } k=4$

Step-by-step explanation:

We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a binomial in the form (x - a), then our remainder will be P(a).

We are dividing:

$(x^2+5x+7)\div(x+k)$

So, a polynomial by a binomial factor.

Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.

We want our remainder to be 3. So, P(a)=P(-k)=3.

Therefore:

$(-k)^2+5(-k)+7=3$

Simplify:

$k^2-5k+7=3$

Solve for k. Subtract 3 from both sides:

$k^2-5k+4=0$

Factor:

$(k-1)(k-4)=0$

Zero Product Property:

$k-1=0\text{ or } k-4=0$

Solve:

$k=1\text{ or } k=4$

So, either of the two expressions:

$(x^2+5x+7)\div(x+1)\text{ or } (x^2+5x+7)\div(x+4)$

Will yield 3 as the remainder.

5 0

3 years ago

Sanya noticed that the temperature was falling at a steady rate of 1.4 degrees every hour after the time that she first checked

kakasveta [241]

Answer:A not b or c or d A!!

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Y = x/3 - 4 Change this equation into y=mx+b form

dusya [7]

Y=3x+4 beacause the slope is m and b is the y-intercept

5 0

3 years ago

Read 2 more answers