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saw5 [17]
3 years ago
13

A) Write a rule for a polynomial with left end up and right end up behavior, and has a y-

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
7 0

Answer:

y = x^2 + 3

Step-by-step explanation:

The equation I put has the left end up, the right end up, and has a y-intercept at (0,3).

If this answer is correct, please make me Brainliest!

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What is the geometric definition of a circle
Mazyrski [523]

Answer:

See explanation

Step-by-step explanation:

A circle is a figure where all of the points on its graph are equidistant from a single, central point. You can define a circle in a plane with the formula (x-h)^2+(y-k)^2=r^2, where h represents the x coordinate of the center, k represents the y coordinate, and r represents the radius. Hope this helps, and feel free to ask for more clarification!

3 0
3 years ago
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Find the y-intercept of each exponential function and order the functions from least to greatest y-intercept.
kvv77 [185]

Answer:

Keep the exact order that is shown

Step-by-step explanation:

The first exponential already has a point at (1, -2) which means the y-intercept would be less than -2, which is the lowest among the three


The second has a y-int at -2

Third has y-int at 2

4 0
2 years ago
A line with slope 3/8 passes through the point -4 0 find another point on the same line​
Mamont248 [21]

Answer:

(0,3/2)

Step-by-step explanation:

y-y1=m(x-x1)

y-0=3/8(x- -4)

y= 3/8x+ 3/2

3/2= y intersept

(0,3/2)= second point

5 0
3 years ago
What are the points for f(x)=3^(x-1)-2
Savatey [412]

Answer:  (0, -1 \frac{2}{3}), (1, -1), (2, 1)

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

Set the exponent equal to zero - <em>that is your anchor point</em>.

Then choose an x-value less than and greater than the anchor point.

\left\begin{array}{c|c|l}&\underline{\quad x\quad }&\underline{\qquad \qquad f(x)\qquad \qquad \qquad \quad }\\\text{less than anchor}&0&3^{0-1}-2=3^{-1}-2=\frac{1}{3}-2 = -1\frac{2}{3}}\\\\anchor\ point&1&3^{1-1}-2=3^{0}-2=1-2 = -1}\\\\\text{greater than anchor}&2&3^{2-1}-2=3^{1}-2=3-2 = 1}\end{array}\right

4 0
3 years ago
What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?
Likurg_2 [28]
Force = mass x exceleration 
1100 x 0.5 = 550
8 0
3 years ago
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