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2 years ago

Help me to understand this

Mathematics

1 answer:

geniusboy [140]2 years ago

4 0

Triangle....
three angles of any triangle have sum equals 180
so
m<A + m<B +m<C = 180 then substitute values and solve for x

I can't see clearly on m<A...is it the value 35? if so
x+35+65 = 180
x+100 = 180
x = 180 -100
x = 80

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A rectangular piece of metal is 20 in longer than it is wide. Squares with sides 4 in long are cut from the four corners and the

Misha Larkins [42]

Answer:

Step-by-step explanation:

This is good! The length of the metal is 20 inches longer than the width. But we fold up 4-inch flaps on each side, so the height of the box is 4 inches, and the length and width will each decrease by 8 inches (4 inches on each side). Since the length and width decrease by the same amount, the length will still be 20 inches longer than the width. So the equations would be:

l = w + 20

vol = l x w x h = 1716 in3

1716 = (w + 20)(w)(4)

1716 = 4w2 + 80w

4w2 + 80w - 1716 = 0

4(w2 + 20w - 429) = 0

4(w + 33)(w - 13) = 0

w + 33 = 0, so w = -33

w - 13 = 0, so w = 13

Since the width can't be negative, the width of the box is 13 and the length is 33 (13 + 20).

But remember, the length and width of the piece of metal are each 8 inches longer than the box, so it was 41 inches by 21 inches.

7 0

2 years ago

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$