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2 years ago

Very sleepy plz help, been doing these all day... question in photo

Mathematics

1 answer:

zzz [600]2 years ago

6 0

Answer:

20 40 and 90 are correct

Step-by-step explanation:

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Write an equation in slope-intercept form for the line with slope -7 and y-intercept 6

Lera25 [3.4K]

Answer:

y=-7x+6

Step-by-step explanation:

slope =m

y intercept =b

y=mx+b

y=-7x+6

6 0

3 years ago

Read 2 more answers

Please help me with this someone please explain and show your full work

Marysya12 [62]

1. I am guessing the triangles split the bottom in half so you will find the area of the triangles and the rectangle and add them together
Triangle=3×4 /2 12/2=6
Triangle 3×4 /2 6
Rectangle: 8×1 8
6+6+8=
12+8=20m^2

2. Triangle area+rectangle area
Rectangle: 22×24 528
Triangle: 8×24 /2 192/2 96
528+96= 624

5 0

3 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,

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