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4 years ago

9

Which function has a y-intercept of 3?

1 answer:

liq [111]4 years ago

3 0

If you have choices, you should list them. If you are just asking about any example, then here are some.

y = a*x + 3 ; the value of a can be any real number.

- 3x + y = 3
2y + 4x = 6

ny = ax + 3*n When you divide through by n, you get
y = a/n * x + 3n/n
y = ax/n + 3

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lara31 [8.8K]

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3 years ago

Evaluate 5 - (4 + 2)2<br> O A. 9<br> O B. -15<br> O C. 5<br> O D.-31

solong [7]

Answer:

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Step-by-step explanation:

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3 years ago

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Art [367]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Transform each polar equation to an equation in rectangular coordinates and identify its shape.

Gemiola [76]

Answer:

(a) $x ^ 2 + y ^ 2 = 6 ^ 2$

(b) $(x-1) ^ 2 + y ^ 2 = 1$

Step-by-step explanation:

Remember that to convert from polar to rectangular coordinates you must use the relationship:

$x = rcos(\theta)$

$y = rsin(\theta)$

$x ^ 2 + y ^ 2 = r ^ 2$

In this case we have the following equations in polar coordinates.

(a) $r = 6$ .

Note that in this equation the radius is constant, it does not depend on $\theta$ .

As $r ^ 2 = x ^ 2 + y ^ 2$

Then we replace the value of the radius in the equation and we have to::

$x ^ 2 + y ^ 2 = 6 ^ 2$

Then $r = 6$ in rectangular coordinates is a circle centered on the point (0,0) and with a constant radius $r = 6$ .

(b) $r = 2cos(\theta)$

The radius is not constant, the radius depends on $\theta$ .

To convert this equation to rectangular coordinates we write

$r = 2cos(\theta)$ <em>Multiply both sides of the equality by </em><em>r</em>.

$r ^ 2 = 2 *rcos(\theta)$ <em>remember that</em> $x = rcos(\theta)$ , <em>then: </em>

$r ^ 2 = 2x$ <em>remember that </em> $x ^ 2 + y ^ 2 = r ^ 2$ , <em>then:</em>

$x ^ 2 + y ^ 2 = 2x$ <em>Simplify the expression. </em>

$x ^ 2 -2x + y ^ 2 = 0$ <em>Complete the square. </em>

$x ^ 2 -2x + 1 + y ^ 2 = 1$

$(x-1) ^ 2 + y ^ 2 = 1$ <em>It is a circle centered on the point (1, 0) and with radio $r=1$ </em>

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3 years ago

Algebra ∠KPL and ∠JPL are a linear pair, m∠KPL=1x+28, and m∠JPL=4x+30.

alexdok [17]

Answer:

well done .good luck.....

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2 years ago