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3 years ago

Subtract (2x²+5) - (4x - 3)

Mathematics

2 answers:

Ivahew [28]3 years ago

3 0

Answer:

x = -1

Step-by-step explanation:

1.1 Pull out like factors :

-2x - 2 = -2 • (x + 1)

2.1 Solve : -2 = 0

x = -1

Tpy6a [65]3 years ago

3 0

Answer: 2x^2 - 4x + 8

Step-by-step explanation:

The first step to solving this equation will be to distribute the negative sign to the 4x and the -3 in parenthesis.

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

We now have the equation:

2x^2 + 5 -4x + 3

Combine like terms:

2x^2 - 4x + 8

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Angle V is 40. Angle V and Y are congruent. The angles of a triangle are a sum of 180. Angle V and Angle Y add up to 80. Therefore angle VWZ is 100

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PLEASE HELP! I WILL MARK TOU AS BRAINLEIST!

san4es73 [151]

Answer:

a.Z(-2,1)

b.Z(1,1)

c.Z(-3,2)

Step-by-step explanation:

z(-2,3)

Imagine this point on a graph.

Translate it down two units :

the x stays -2, by going down the y decreases 2 so 3-2=1

Z(-2,1)

Translate Right three units : I'm assuming that we use the answer from the first translation

Z(-2,1)

The y doesn't change this time the x increases 3 since we're moving to the right.

Z(1,1)

Translate up 1 and left 4:

Z(1,1)

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3 years ago

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The sum of 3 interior angles of a quadrilateral is 262°. Find the measure of the fourth interior angle.

Vilka [71]

The four interior angles of a quadrilateral always add to 360°, so the answer is 98

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The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

and volume,

$V=\frac{1}{3}\pi r^2 h$

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

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