What is the sum of the polynomials?

The answer is 4 you just have to figure out what x is and fill that in and whatever that is it should give u the answer as 4

Marat540 [252]

Hello. 10x-3-2x=4 ; 10x-2x=4+3 ; 8x=7 ; x=7/8. I hope to have helped you

5 0

3 years ago

Solve for x: x+10 = 4 2 -

ELEN [110]

Simplify
3
x
2
−
x
3
x
2
-
x
.
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x
2
=
x
10
−
4
5
x
2
=
x
10
-
4
5
Move all terms containing
x
x
to the left side of the equation.
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2
x
5
=
−
4
5
2
x
5
=
-
4
5
Since the expression on each side of the equation has the same denominator, the numerators must be equal.
2
x
=
−
4
2
x
=
-
4
Divide each term by
2
2
and simplify.
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x
=
−
2

6 0

3 years ago

Read 2 more answers

How many years would it take for a person in the united states to eat 855 pounds of apples

never [62]

It would take 30 years

4 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

Consider the following system:

Goryan [66]

You did not include the choices. However, I answered one that just included them. I've included the possible answers below and then the correct answers.

A multiple of Equation 1.
B. The sum of Equation 1 and Equation 2
C. An equation that replaces only the coefficient of x with the sum of the coefficients of x in Equation 1 and Equation 2.
D. An equation that replaces only the coefficient of y with the sum of the coefficients of y in Equation 1 and Equation 2.
E. The sum of a multiple of Equation 1 and Equation 2.

A, B and E.

Adding and multiplying the terms allow them to keep working. However, you must make sure that each variable is changed each time. Not just one as in C and D.

4 0

3 years ago

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