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svetlana [45]
3 years ago
6

Draw a line representing the rise and a line representing the run of the line State the slope of the line in simplest form

Mathematics
1 answer:
amid [387]3 years ago
8 0
Slope is -5/3
I hope this helps :)

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The average value of y=v(x) equals 4 for 1≤x≤6, and equals 5 for 6≤x≤8. what is the average value of v(x) for 1≤x≤8 ?
Tomtit [17]
The average value over the interval is the area under the curve divided by the width of the interval.

Area = 4*(6 -1) +5(8 -6) = 30
Width = 8 - 1 = 7

Average value = 30/7 = 4 2/7

7 0
3 years ago
I don’t know what to do on here it’s really hard and I need help for the answer
Alla [95]
Ok so our original fraction is:
\frac{36 {x}^{12} }{6 {x}^{9} }
To simplify this fraction, look for instances where the values on the top and bottom can be reduced:

For example, 36 over 6 is the same as 6 over 1, so we can simplify the fraction so it is:
\frac{6 {x}^{12} }{ {x}^{9} }
We can also eliminate the denominator by dividing the nominator by x^9 so:
\frac{6 {x}^{12} }{ {x}^{9} }  \div  \frac{ {x}^{9} }{ {x}^{9} }
= 6 {x}^{3}
And that is the simplified answer of the fraction

Hope this helped
5 0
3 years ago
A fence is 4 feet tall. The fence has a shadow that is 5 feet long. Paul is 6 feet tall, and he is standing next to the fence. U
adoni [48]

Answer:

7.5 feet

Step-by-step explanation:

4/5=6/x

cross multiply

4x=30

x=7.5

5 0
3 years ago
Read 2 more answers
Solve 4/x-4=x/x-4-4/3 for x and determine if the solution is extraneous or not
4vir4ik [10]

<u>Answer:</u>

x = 4 (extraneous solution)

<u>Step-by-step explanation:</u>

\frac { 4 } { x - 4 } = \frac { x } { x - 4 } - \frac { 4 } { 3 } \\ \frac { 4 } { x - 4 } - \frac { x } { x - 4 } = - \frac { 4 } { 3 } \\ \frac { 4 - x } { x - 4 } = - \frac { 4 } { 3 } \\ 3 ( 4 - x ) = - 4 ( x - 4 ) \\ 1 2 - 3 x = - 4 x + 1 6 \\ 4 x - 3 x = 1 6 - 1 2 \\ x = 4 \\

This solution is extraneous. Reason being that even if it can be solved algebraically, it is still not a valid solution because if we substitute back x=4, we will get two fractions with zero denominator which would be undefined.

7 0
3 years ago
Verify the trigonometric identities
snow_lady [41]
1)

here, we do the left-hand-side

\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2&#10;\\\\\\\&#10;[sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)]&#10;\\\\\\&#10;2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2

2)

here we also do the left-hand-side

\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x)&#10;\\\\\\&#10;\cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)}&#10;\\\\\\&#10;\cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)

3)

here, we do the right-hand-side

\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}&#10;\\\\\\&#10;\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}}&#10;\\\\\\&#10;\cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}
8 0
3 years ago
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