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borishaifa [10]

3 years ago

6

Subtract 2x + 1 from 3x^2 - 6x + 2

1 answer:

mash [69]3 years ago

5 0

Match variables to subtract properly:
-6x-2x=-8x
2-1=1

So we are left with: 3x^2-8x+1

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Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6

kenny6666 [7]

Answer:

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

Step-by-step explanation:

Given the simultaneous equations

$y=9-x$

$y\:=\:2x^2\:+\:4x\:+\:6$

Subtract the equations

$y=9-x$

$-$

$\underline{y=2x^2+4x+6}$

$y-y=9-x-\left(2x^2+4x+6\right)$

$\mathrm{Refine}$

$x\left(2x+5\right)=3$

$\mathrm{Solve\:}\:x\left(2x+5\right)=3$

$2x^2+5x=3$ ∵ $\mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x$

$\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}$

$2x^2+5x-3=3-3$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{Quadratic\:Equation\:Formula:}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\$

$x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$=\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{4}$

$=\frac{-5+7}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Similarly,

$x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{1}{2},\:x=-3$

$\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12$

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

3 0

4 years ago

Katie's swim team is having an end-of-season pizza party. When the party started, there were 50 slices of pizza. Now there are 2

ss7ja [257]

Answer:

i think its c im not sure so it might be wrong im so sorry if its wrong

Step-by-step explanation:

6 0

3 years ago

Find the product of (x 5)(x − 5). X2 − 10x 25 x2 10x 25 x2 − 25 x2 25.

den301095 [7]

Product of two polynomials can be done by distribution of multiplication over addition. The product of $(x+5)(x-5) = x^2 - 25$

<h3>What is distributive property of multiplication over addition?</h3>

Suppose a, b and c are three numbers. Then we have:

$a(b + c) = a\times b + a\times c$

(a(b+c) means a multiplied to (b+c). The sign of multiplication is usually hidden when using symbols and both quantities which are in multiplication are written together without space)

The product result of the given expression is obtained as:

$(x+5)(x-5) = x(x - 5) + 5(x - 5) \\\\(x+5)(x-5) = x^{1+1} - 5x + 5x - 25 = x^2 - 25$

Thus,

The product of $(x+5)(x-5) = x^2 - 25$

Learn more about polynomial product here;

brainly.com/question/9106484

4 0

2 years ago

Real numbers consist of

gregori [183]

Answer:

a

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

How much greater is the area of a circle with a radius of 6.1 inches than the area of a square with a side length of 4 inches?

umka2103 [35]

Answer:

The area of the circle is 100.8394 square inches greater than the area of the square.

Step-by-step explanation:

Given:

The radius of the circle 6.1 inches

side of the square= 4 inches

Finding the area of the circle

Area of the circle= $\pi r^2$

= $\pi *6.1*6.1$

$=3.14*6.1*6.1$

= 116.8394 square inches

Area of the square= side*side

$=4*4\\ =16 square inches$

$116.8394-16=100.8394$

The area of the circle is 100.8394 square inches greater than the area of the square.

4 0

3 years ago