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Maru [420]
3 years ago
5

8u+9v+3u+5v simplify.​

Mathematics
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer: 11 u + 14 v

Step-by-step explanation:

Just add the like terms together.

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PLEASE HELP <br>I need the answers for all of the ones that I haven't done please. Thank you!
Rudik [331]
#2.

a) 1/4m + 3 = 2

1/4m = -1

m = -4

b) 4y - 3(y + 8) = 12

4y - 3y + 24 = 12

y = -12



For #10, letter b. Do the same thing you did with letter a.
The equation is already given.
You have your y-intercept: -8, and your slope: 3/1.

These two create two different points you can add to your data table.

(0, -8) and (1, 3)

y:-8, 3
x: 0, 1

6 0
4 years ago
30 POINTS<br><br> why are 3/4 and 6/8 equivalent
Flauer [41]

Step-by-step explanation: In order to decide whether the given fractions are equivalent, first write each fraction in lowest terms. Remember, a fraction is written in lowest terms if the greatest common factor of the numerator and the denominator is 1.

For our first fraction which is 3/4, the greatest common factor of 3 and 4 is 1, the fraction is already in lowest terms.

For our second fraction however, since the greatest common factor of 6 and 8 is 2, we must divide both the numerator and denominator by 2 which gives us 3/4.

3/4 = 3/4 which means these fractions are equivalent.

3/4 and 6/8 are equivalent because if you reduce 6/8 in lowest terms, you end up with the fraction 3/4.

3 0
4 years ago
Please help I'm not good with word problems ​
rusak2 [61]

Answer:

7 5/8

Step-by-step explanation:

5+2= 7 3/8+2/8=5/8 7+5/8=7 5/8

6 0
3 years ago
Read 2 more answers
Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A
mash [69]

Answer:

Step-by-step explanation:

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA

we take the LHS so here goes,

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\

since , sin^2A+cos^2A=1

the identity becomes,

(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\

now, we know,

sin^2A=1-cos^2A and cos^2A=1-sin^2A

the identity becomes,

(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\

(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})

sin A and cos A cancel out it becomes zero

\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\

Splitting the denominator the identity becomes

\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A}  \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\

Hence,

tanAsecA-cotAcosecA

3 0
3 years ago
Can someone solve this please<br><br>it is solving equations in algebra: x-6=4​
Sholpan [36]
The answer to this equation is 10 = x
7 0
3 years ago
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