What is the 4(4x)+(x) equal!?

The angle of depression from D to F measures 40°. If DE = 100 m, find DF. Round your answer to the nearest tenth.

leva [86]

You need to draw the triangle DEF with a right angle at E. Angle D = 40 and DE = 100 m. Once you have marked up your triangle, you can use cosine to find DF.

7 0

3 years ago

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2x^3-x^2-3x=210 the answer is 5 but I want to know why.

mel-nik [20]

Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

-------------------------------------------------------------------------

$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

-----------------------------------------------------------------------

$9x^{2}$ $-3x$ $-210$

--------------------------------------------------------------------------

$42x$ $-210$

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

4 0

3 years ago

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On Monday the temperature was -5°F. How many degrees below zero was the temperature

stira [4]

The temperature was 5 degrees below 0

7 0

3 years ago

Write a proof for the biconditional using the inverse and contrapositive.

MariettaO [177]

I don’t really understand the question at hand but let me try to help

X is equal to 4 we know this because because the sign in between the number means greater than or equal to so 3 x 4 equals 12 and that makes the equation on right true and on the left x is also equal to for which also makes the statement on the right true as well so I think X = 4

I hope I was able to help and If not I’m sorry

7 0

3 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

3 years ago