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4 years ago

Find the y-intercept of the function represented by the graph

Mathematics

1 answer:

dangina [55]4 years ago

4 0

Answer:

Step-by-step explanation:

I THINK

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What % of 400 = 193

saveliy_v [14]

The best way to find your answer is to divide 193 by 400 to get .4825 and to make sure if it is the answer you take 400 and multiply .4825 and it will be 193 so .4825 = 48.25%

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3 years ago

Y^-1 dy +ye^(cosx) sinxdx=0

olga nikolaevna [1]

Please see the answer here

http://www.wolframalpha.com/input/?i=y%5E-1%20dy%20%2Bye%5E%28cosx%29%20sinxdx%3D0

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3 years ago

Which is true 10.59>10.60<br> 10.590=10.59<br> 10.43<10.24 <br> 9.0005=9.0050

mart [117]

I think the second i didnt make it on paper but its the one that more convence me

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3 years ago

Write a parallel and perpendicular equation that passes through the point (6,-2)?

Reptile [31]

Answer:

y=x-8 and y=-x+4

Step-by-step explanation:

The slopes of the lines can be anything, as long as they are the opposite reciprocals of each other. Then you can plug in the point for x and y to solve for b in both lines (for y=mx+b where m is the slope)

3 0

3 years ago

Simplify the expression csc(-x)/1+tan^2x)

Charra [1.4K]

Assuming ya meant $\frac{csc(-x)}{1+tan^2(x)}$

to slimplify, we use a variation of the pythagorean identity and a decomposition into the sin and cos

for the pythaogreaon identity
$cos^2(x)+sin^2(x)=1$
divide both sides by $cos^2(x)$
$1+tan^2(x)=sec^2(x)$ since $\frac{sin(x)}{cos(x)}=tan(x)$
subsitute
$\frac{csc(x)}{sec^2(x)}$

recall that $csc(x)=\frac{1}{cos(x)}$
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore $csc(-x)=\frac{1}{cos(-x)}=\frac{1}{cos(x)}=csc(x)$
so we get

$\frac{csc(x)}{sec^2(x)}$
decompose them into $\frac{1}{cos(x)}$ and $\frac{1}{sin^2(x)}$ to get $\frac{\frac{1}{cos(x)}}{\frac{1}{sin^2(x)}}$
multiply by $\frac{sin^2(x)}{sin^2(x)}$ to get
$\frac{sin^2(x)}{cos(x)}$
we can furthur simlify to get
$(\frac{sin(x)}{cos(x)})(sin(x))=tan(x)sin(x)$
the expression simplifies to tan(x)sin(x)

5 0

3 years ago

Read 2 more answers