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Reil [10]
3 years ago
8

Solving Linear Equations: Variables on Both Sides I really need this please!

Mathematics
1 answer:
damaskus [11]3 years ago
4 0
Answer: x=-2/3

Explanation: Heres what my calculator says!
After you divide it will give you -2/3 :)

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What are the zeros of f(x)=x^2-10+25
BartSMP [9]
Answer:
-5

Explanation:
Because it’s right
7 0
3 years ago
Evaluate b-(-1/8)+c where b=2 and c=-7/4​
tatiyna

Answer:

3/8

Step-by-step explanation:

Evaluate b - -1/8 + c where b = 2 and c = -7/4:

b - (-1)/8 + c = 2 - (-1)/8 - 7/4

Put 2 + 1/8 - 7/4 over the common denominator 8. 2 + 1/8 - 7/4 = (8×2)/8 + 1/8 + (2 (-7))/8:

(8×2)/8 + 1/8 - (7×2)/8

8×2 = 16:

16/8 + 1/8 - (7×2)/8

2 (-7) = -14:

16/8 + 1/8 + (-14)/8

16/8 + 1/8 - 14/8 = (16 + 1 - 14)/8:

(16 + 1 - 14)/8

16 + 1 = 17:

(17 - 14)/8

| 1 | 7

- | 1 | 4

| 0 | 3:

Answer:  3/8

8 0
3 years ago
What is g(x)?<br> A g(x)=-x2<br> B g(x)=-2x<br> C g(x)= -x<br> D g(x)= -|x|
boyakko [2]

Answer:

Should be A

Hope this helped!

8 0
3 years ago
Read 2 more answers
Guys please help! In triangle CDE, the measure of angle C is 90 degrees, side CD is 15 and DE is 3 more than CE. Find the length
AURORKA [14]
You can calculate it using the law of cosines: c^2=a^2+b^2-2*a*b*cos(C)

your triangle is
CD=15=a
CE=?=b
DE=CE+3=b+3=c
and C=90°

-> insert those values, with c substituted with b+3 to remove c

c^2=a^2+b^2-2*a*b*cos(C)
(b+3)^2=15^2+b^2-2*15*b*cos(90)
cos(90)=0->
(b+3)^2=15^2+b^2
b^2+2*3*b+3^2=225+b^2
6b+9=225
6b=216
b=36=CE

DE=CE+3=36+3=39
4 0
3 years ago
HELP ME ASAP!!!!!!!!
coldgirl [10]

Step-by-step explanation:

a. 6 £/hr

b. 6.8 £/hr.....is this right?

8 0
3 years ago
Read 2 more answers
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