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solmaris [256]
3 years ago
7

Find the slope of the line that passes through (9,3) and (6,4)​

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
6 0
Slope of the line that passes through is -1/3x+6
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5.8 Sq ft= _____ Sq in? Please give me the correct answer. I need help. Give me a whole number or decimal, thanks!
nekit [7.7K]
68 sq in. You multiply 5 by 12 ( amount of inches in a foot) and get 60, then add the extra eight inches(these inches are from 5.8). Hope this helps :)
3 0
3 years ago
Which represents a perfect cube? 6 times 6 times 6 6 + 6 + 6 3 times 3 times 3 times 3 3 + 3 + 3 + 3 + 3
mojhsa [17]

Answer:

6 times 6 times 6

Step-by-step explanation:

6 x 6 = 6²

Therefore 6 x 6 x 6 = 6³

4 0
4 years ago
Read 2 more answers
Mark true or false to indicate whether the ratios are equivalent or not
Ann [662]

Answer:

1) False

2) True

3) True

4) False

Step-by-step explanation:

Divide second numerator by the first numerator

Do this to the denominators as well

Compare the numbers and if they are the same then they are equal to eachother

8 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
How to solve y=|4x+2| in absolute value
meriva
Since this is this equation has an absolute value than anything in the absolute value is positive.
y=|4x+2|
y=|6x|
y=6x
7 0
3 years ago
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