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4 years ago

5

15 POINT QUESTION Which polynomial is a quintic trinomial?

1 answer:

kozerog [31]4 years ago

8 0

A quintic polynomial means a polynomials who's highest exponent value is 5. And a trinomial is a polynomial that has exactly three terms. So the answer is C

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16% of a number is 60. What is that number?<br><br> A375<br> B 432<br> C 456<br> D753

Vlad1618 [11]

Answer:

Its A: 375

Because, if you take 26.6 you have to round it. So, the 6 behind the decimal is over 5 so make the other 6 in front of the decimal a 7. So, the 7 will add 1 to the 2 to make it 3. And i forgot how to get the 5, but trust me dis is right. :3

4 0

3 years ago

5 times blank equal 550

FromTheMoon [43]

110. 550 ÷ 5 = 110.
5 x 110 = 550.

3 0

3 years ago

What is the value of the expression 2 1/2 x + 3 2/3 y ( if x = 4 and y= 2)

vivado [14]

Answer:

17 1/3

Step-by-step explanation:

First you place 4 in for x:

2 1/2(4)=10

Then place 2 in for y:

3 2/3(2)=7 1/3

Finally add both together:

10+7 1/3= 17 1/3

4 0

3 years ago

I need help with this one

Tomtit [17]

Answer:

mean > median

Step-by-step explanation:

The graph is skewed right since the data is to the left.

When the graph is skewed right, the median is usually less than the mean

This means the mean > median

6 0

3 years ago

Read 2 more answers

Find the equation of a circle that has a diameter with the endpoints given by the points A(-4,9) and B (- 2, - 3) )

Aleksandr-060686 [28]

The equation of the circle is $(x+3)^{2}+(y-3)^{2}=37$

Explanation:

Given that the endpoints of the circle are A(-4,9) and B(-2,-3)

We need to determine the equation of the circle.

<u>Center:</u>

The center of the circle can be determined using the midpoint formula,

$Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substituting the coordinates A(-4,9) and B(-2,-3), we get,

$Center=(\frac{-4-2}{2},\frac{9-3}{2})$

$Center=(\frac{-6}{2},\frac{6}{2})$

$Center=(-3,3)$

Thus, the center of the circle is (-3,3)

<u>Radius:</u>

The radius of the circle can be determined using the distance formula,

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Substituting the center (-3,3) and the endpoint (-4,9), we get,

$r=\sqrt{\left(-4+3\right)^{2}+\left(9-3\right)^{2}}$

$r=\sqrt{\left(-1\right)^{2}+\left(6\right)^{2}}$

$r=\sqrt{1+36}$

$r=\sqrt{37}$

Thus, the radius of the circle is $\sqrt{37}$

<u>Equation of the circle:</u>

The standard form of the equation of the circle is given by

$(x-a)^{2}+(y-b)^{2}=r^{2}$

where (a,b) is the center and r is the radius.

Substituting the values, we have,

$(x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}$

$(x+3)^{2}+(y-3)^{2}=37$

Thus, the equation of the circle is $(x+3)^{2}+(y-3)^{2}=37$

7 0

3 years ago