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3 years ago

Help me with this problem plzzzz

Mathematics

2 answers:

xeze [42]3 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

∠7 = 137° ( vertically opposite angle )

∠8 = 180° - ∠7 = 180° - 137° = 43° ( straight angle )

∠6 = ∠8 = 43° ( vertically opposite angles )

fiasKO [112]3 years ago

5 0

Answer:

7= 137

6= 43

8=43

Step-by-step explanation:

well 7 would be the same angle as 137

and 6 and 8 are the same angle

so u already know that

7= 137

now to find 6 0r 8 u just need to subtract 137 from 180 to get

6= 43

8= 43

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Simplify the expression state any excluded values 2a^2-4a+2 --------------- 3a^2-3

chubhunter [2.5K]

Answer:

The simplified form is $\dfrac{2(x-1)}{3(x+1)}$ .

$x =1$ is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

$\dfrac{2a^2-4a+2}{3a^2-3}$

Let us simplify the numerator and denominator separately.

The numerator is given as $2a^2-4a+2$

2 is a common factor in all the three terms. So, we factor it out. This gives,

$=2(a^2-2a+1)$

Now, $a^2-2a+1=(a-1)(a-1)$

Therefore, the numerator becomes $2(a-1)(a-1)$

The denominator is given as: $3a^2-3$

Factoring out 3, we get

$3(a^2-1)$

Now, $a^2-1$ is of the form $a^2-b^2=(a-b)(a+b)$

So, $a^2-1=(a-1)(a+1)$

Therefore, the denominator becomes $3(a-1)(a+1)$

Now, the given expression is simplified to:

$\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}$

There is $(x-1)$ in the numerator and denominator. We can cancel them only if $x\ne1$ as for $x=1$ , the given expression is undefined.

Now, cancelling the like terms considering $x\ne1$ , we get:

$\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}$

Therefore, the simplified form is $\dfrac{2(x-1)}{3(x+1)}$

The simplification is true only if $x\ne1$ . So, $x =1$ is the excluded value for the given expression.

8 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:

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