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3 years ago

Is 10x is polynomial or not?

Mathematics

1 answer:

Inga [223]3 years ago

6 0

Answer:

Yes, 10x is a polynomial.

Step-by-step explanation :

In particular, for an expression to be a polynomial term, it must contain no square roots of variables, no fractional or negative powers on the variables, and no variables in the denominators of any fractions. That's why 10x is a polynomial because it obeys all the rules.

Hence, 10x is a polynomial.

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Answer:

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Jeff is 5 feet tall and casts a shadow 30 inches long. How tall is Marisol if her shadow is 24 inches long?

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Read 2 more answers

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

For the first parenthesis:

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

For the second parenthesis:

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$

8 0

3 years ago

Oz= 1 c please help with this question

MaRussiya [10]

The answer is 8

hope this helped :)
alisa202

8 0

3 years ago

Someone please help marking brainliest

Luda [366]

Answer: (1, 4)

Explanation: When using the method of elimination, the goal is to eliminate a variable by either adding or subtracting the 2 equations. For this question, you can choose either to eliminate X or Y. I’ll eliminate X as an example:

In order to eliminate a variable, the same variable in both equations must have the same coefficient.

(1) 3x+y=7
(2) 2x+5y=22

Multiply (1) by 2:

(3) 6x+2y=14

Multiply (2) by 3:

(4) 6x+15y=66

Now that X in both equations has the same coefficient of 6, you can subtract the two equations to officially eliminate the variable and solve for Y:

Subtract (4) from (3):

-13y=-52
y=4

Now that you have the value of Y, substitute that into either one of the equations to get X. I’ll use the first equation as an example:

3x+(4)=7
3x=3
x=1

Therefore, the point of intersection is (1, 4).

Hope this helps シ

3 0

3 years ago