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4 years ago

HELPPPP PLZZ QUICK I WILL GIVE BRSINLEST

Mathematics

2 answers:

olasank [31]4 years ago

7 0

A refraction cup is in the shape of a half circle. They are flat on the top and bottom with one flat edge.

If you were to put the two flat edges together it would make a 3 dimensional cylinder

jok3333 [9.3K]4 years ago

3 0

A refraction cup that is the shape of the semicircle will create a sphere if you put two of them together.

Hope this helps:)

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If the length of AC equals 50, what is the length of the midsegment DE?

balu736 [363]

Answer:

Step-by-step explanation:

The mid-segment of the triangle is always parallel to the base and is half the length of the base

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3 years ago

The following data of students weights (in kg) is collected 65 48 52 55 62 58 47 53 65 71 54 62 51 49 54 60 68 53 57 62 59

tatuchka [14]

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Step-by-step explanation:

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3 years ago

If a polynomial function f(x) has roots 3+Sqrt(5) and -6, what must be a factor of f(x)

neonofarm [45]

Answer:

The two factors of the given polynomial f(x) are (x - 3 -√5) and (x +6) .

Step-by-step explanation:

Here, the roots of the given polynomial f(x) are

Root 1 = 3 + √5

and Root 2 = (-6)

Now,as we know that if x = a is the root of the any given polynomial p(x), then the expression (x-a) is the FACTOR of the polynomial p(x).

Now, here root 1 : x = 3 + √5

So, x - (3 + √5 ) is the FACTOR of the f(x)

or, ( x - 3 -√5) is the first factor of f(x).

Also, here root 2 : x = (-6)

So, x - (-6 ) = x + 6 is the FACTOR of the f(x)

or, ( x +6 ) is the second factor of f(x).

Hence, the two factors of the given polynomial f(x)

are ( x - 3 -√5) and ( x +6 ) .

5 0

4 years ago

Checking the Mean Value Theorem:<br><br>Number 3

mrs_skeptik [129]

$\bf f(x)=x+\cfrac{1}{x}\qquad \left[\frac{1}{2},2 \right]\\\\
-----------------------------\\\\
\cfrac{df}{dx}=1+\left(-1x^{-2} \right)\implies \cfrac{df}{dx}=1-\cfrac{1}{x^2}
\\\\\\
f'(c)=1-\cfrac{1}{c^2}\quad \quad 1-\cfrac{1}{c^2}=\cfrac{f(2)-f\left( \frac{1}{2} \right)}{2-\frac{1}{2}}
\\\\\\
1-\cfrac{1}{c^2}=\cfrac{\frac{5}{2}-\frac{5}{2}}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=\cfrac{0}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=0
\\\\\\
1=\cfrac{1}{c^2}\implies c^2=1\implies c=\pm\sqrt{1}\implies c=\pm 1$

there's a quick graph below of the bounds and the tangent at "c"

not happening -2 or 2 will have a tangent parallel to a,b, needless to say -2 is out of the range [a,b] anyway, so the only value is really 1, on the positive 1st quadrant

6 0

3 years ago

The function y = -x - 1 and y = ½x + k intersect at the point (-2,1). Find the value of k.

horrorfan [7]

Answer:

k = 2

Step-by-step explanation:

Fill in the given point values and solve for k.

y = (1/2)x + k

1 = (1/2)(-2) +k . . . . . . for x = -2, y = 1

1 = -1 +k . . . . simplify; next, add 1

2 = k

5 0

4 years ago