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3 years ago

Solve for x; 2x+3=11?

Mathematics

2 answers:

Blababa [14]3 years ago

8 0

Answer:

2x+3=11

2x=11-3

x=8/2

x=4........

MaRussiya [10]3 years ago

5 0

Answer: $x = 4$

Step-by-step explanation:

Okay, so we start with our equation.

$2x + 3 = 11$ .

Then we need to get x alone. To do that we subtract the 3 on the side with the x.

But what we do to one side, we do to the other. So we subtract 3 from the 11 as well

2x + 3 = 11

-3 -3

2x = 8

Then we divide by 2 on both sides to get X.

$\frac{2x}{2} =\frac{8}{2}$

$\frac{2x}{2} = x$

$\frac{8}{2} = 4$

$x = 4$ .

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Add: -5g^2+(-9g^2-8)

Amiraneli [1.4K]

Answer:

$\boxed{-14g^2 - 8}$

Step-by-step explanation:

$-5g^2 + (-9g^2 - 8)$

$= - 5g^2 - 9g^2 - 8$

$= -14g^2 - 8$

3 0

3 years ago

Adam is building a computer desk with a separate compartment for the computer. The compartment for the computer is a rectangular

Alik [6]

Answer:

The width of the computer compartment is 9 inches

The length of the computer compartment is 33 inches

The height of the computer compartment is 27 inches

Step-by-step explanation:

The given data on the rectangular prism compartment are;

The volume of the rectangular prism, V = 8019 cubic inches

The length of the compartment, l = 24 inches + The width, w

The height of the compartment, h = 18 inches + The width, w

Therefore, we have;

l = 24 + w

h = 18 + w

V = l × h × w

∴ V = (24 + w) × (18 + w) × w = w³ + 42·w² + 432·w = 8019

w³ + 42·w² + 432·w - 8019 = 0

By graphing the above function with MS Excel, we have one of the solution is w = 9

∴ (w - 9) is a factor of w³ + 42·w² + 432·w - 8019 = 0

Dividing, we get;

w² + 51·w + 891

w³ + 42·w² + 432·w - 8019 = 0

w³ - 9·w²

51·w² - 459·w

891·w

891·w + 8019

Therefore,

w³ + 42·w² + 432·w - 8019 = 0

w³ + 42·w² + 432·w - 8019 = (x - 9)·(w² + 51·w + 891) = 0

The determinant of the factor, w² + 51·w + 891 = 51² - 4×1×891 = -963, therefore, the it has complex roots

Therefore, the real solution of (x - 9)·(w² + 51·w + 891) = 0 is w = 9

The width of the computer compartment, w = 9 inches

The length of the computer compartment, l = 24 inches + 9 inches = 33 inches

The height of the computer compartment, h = 18 inches + 9 inches = 27 inches.

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3 years ago

Jeanie wants to hang a circular mirror in a frame in her front wall. Determine the area of the largest mirror Jeanie can han if

AnnyKZ [126]

56=2(pi)(r)
r=8.9127
pi(8.9127)^2=249.56cm^2
≈245cm^2 (3s.f.)

3 0

3 years ago

8 squared by 6 ÷ 4 squared by 3 =

loris [4]

8^6 ÷ 4^3

8^6 = 262,144

4^3 = 64

So, 262,144 ÷ 64 = 4,096.

Done!

3 0

3 years ago

Read 2 more answers

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago

Solve for x; 2x+3=11?​

Solve for x; 2x+3=11?