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BabaBlast [244]
3 years ago
7

Solve |2x + 2| = 10.

Mathematics
2 answers:
maks197457 [2]3 years ago
4 0
X=4
(2x4 + 2) =10
(8 + 2) = 10
Usimov [2.4K]3 years ago
3 0

Answer:

Hello!

When solving the equation |2x + 2| = 10 , your answer would be...

x=4,-6

Step-by-step explanation:

Find all solutions for x by changing the <u>absolute value</u> into the positive and negative components

Hope this helps!!

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If AD=6 and AB=24, find AC
ruslelena [56]
When a perpendicular is dropped from the right-angle (C) to the opposite side AB, the metric relations apply:
BD*BA=a^2 ..........................(1)
AD*AB=b^2...........................(2)
BD*DA=DC^2........................(3)

Given AD=6, AB=24, using metric relation (2) above, we have
b^2=6*24=144
=>
b=sqrt(144)=12

By the way, we conclude that this is a 30-60-90 triangle because b/AB=(1/2)=sin(B) => B=30 degrees.

Answer: b=12
6 0
3 years ago
At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not
Nikitich [7]
If f(x, y, z) = c represent a family of surfaces for different values of the constant c. The gradient of the function f defined as \nabla f is a vector normal to the surface f(x, y, z) = c.

Given <span>the paraboloid

y = x^2 + z^2.

We can rewrite it as a scalar value function f as follows:

f(x,y,z)=x^2-y+z^2=0

The normal to the </span><span>paraboloid at any point is given by:

\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\  \\ =2xi-j+2zk

Also, the normal to the given plane 3x + 2y + 7z = 2 is given by:

3i+2j+7k

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</span>
2x=3\Rightarrow x= \frac{3}{2}  \\  \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}

Since, -1 = 2 is not possible, therefore there exist no such point <span>on the paraboloid y = x^2 + z^2 such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2</span>.
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4 years ago
​Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.
Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

x^{3}+7 x^{2}+12 x=0

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

x^{3}+7 x^{2}+12 x=0

x\left(x^{2}+7 x+12\right)=0   ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

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By substituting the values of a,b and c in the quadratic equation we get;

\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}

<em><u>Therefore, the two roots are:</u></em>

\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}

And,

\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

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