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3 years ago

What is the y-intercept for the function?

Mathematics

2 answers:

Serjik [45]3 years ago

7 0

Answer:

The y -intercept of a graph is the point where the graph crosses the y -axis. (Because a function must pass the vertical line test , a function can have at most one y -intercept . ) The y -intercept is often referred to with just the y -value.

Step-by-step explanation:

Y - intercept represents the position of a point on y-axis or when a line passes through y-axis, it, actually, passes through a point on y-axis. And that point is called the y intercept. It is usually represented by and as a point, it is represented as.

It's called the "y intercept" and it's the y value of the point where the line intersects the y- axis. For this line, the y-intercept is "negative 1." You can find the y-intercept by looking at the graph and seeing which point crosses the y axis. This point will always have an x coordinate of zero.

Hoped this helped! :D

bagirrra123 [75]3 years ago

7 0

Y intercept is 8.5 for this function

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After a 90% reduction, you purchase a new soft drink machine on sale for 56$. What was the original price of the soft drink mach

ANTONII [103]

56 is ten percent of the original price.
So 100 percent of the original price would be $560

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3 years ago

Use a graph in a (-2π, 2π, π/2) by (-3, 3, 1) viewing rectangle to complete the identity.

yaroslaw [1]

First, notice that:

$2\tan (\frac{x}{2})=2\cdot(\pm\sqrt[]{\frac{1-cosx}{1+\cos x})}$

And in the denominator we have:

$1+\tan ^2(\frac{x}{2})=1+\frac{1-\cos x}{1+\cos x}=\frac{1+cosx+1-\cos x}{1+cosx}=\frac{2}{1+\cos x}$

then, we have on the original expression:

$\begin{gathered} \frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}=\frac{2\cdot\pm\sqrt[]{\frac{1-\cos x}{1+cosx}}}{\frac{2}{1+\cos x}}=\frac{2\cdot(\pm\sqrt[]{1-cosx})\cdot(1+\cos x)}{2\cdot(\sqrt[]{1+cosx})} \\ =(\sqrt[]{1-\cos x})\cdot(\sqrt[]{1+\cos x})=\sqrt[]{(1-\cos x)(1+\cos x)} \\ =\sqrt[]{1-\cos^2x}=\sqrt[]{\sin^2x}=\sin x \end{gathered}$

therefore, the identity equals to sinx

8 0

1 year ago

I need help! ASAP!! please and thank you

Neko [114]

Answer: The walking path around it.

Step-by-step explanation:

Remember the Pythagorean's theorem, for a triangle rectangle with catheti A and B, and with hypotenuse H, we have:

A^2 + B^2 = H^2

H = √(A^2 + B^2)

Here, A + B is the total distance for the path, while the hypotenuse is the distance for the bridge

Here we can see that:

A = 1170 ft

B = 520 ft

Then:

H = √( (520 ft)^2 + (1170 ft)^2) = 1280.4 ft

Now, let's compute the costs.

For the bridge, we know that each foot costs $11, then for 1280.4 ft the cost is:

Cost of the bridge = (1280.4)*$11 = $14,084.4

And for the walking path, the cost is $6 per foot, then the total cost of the path is:

Cost of the path = (520 ft + 1170 ft)*$6 = $10,140

We know that the bridge is preferred if it is within the range of $1500 for the path's cost.

This range is:

($10,140 - $1,500, $10,140 + $1,500) = ($8,640, $11,640)

Here we can see that the cost of the bridge does not belong to this range, (is higher) so the option we should recommend is the walking path around.

8 0

3 years ago

Explain how to write 50,000 using exponents

WINSTONCH [101]

<u /> $50.000\to \boxed{ 5* 10^4}$

4 0

3 years ago

Read 2 more answers

What is the difference? StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction Sta

katovenus [111]

Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

$\frac{x}{x^2-16}-\frac{3}{x-4}$

To solve the given expression

$\frac{x}{x^2-16}-\frac{3}{x-4}$

$=\frac{x}{x^2-4^2}-\frac{3}{x-4}$

$=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}$ ( using the property $a^2-b^2=(a+b)(a-b)$ )

$=\frac{x-3(x+4)}{(x+4)(x-4)}$

$=\frac{x-3x-12}{(x+4)(x-4)}$ ( by using distributive property )

$=\frac{-2x-12}{(x+4)(x-4)}$

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

$\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

5 0

4 years ago

Read 2 more answers