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3 years ago

Simplify 1/4y +1 1/2+ 2 -1 3/4y -12

Mathematics

2 answers:

sladkih [1.3K]3 years ago

7 0

=-3y-\frac{9}{2}

Hope this Helped!

;D

wel3 years ago

6 0

Based on the equation you gave me, this should be your answer.

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The 5th of 9 consecutive whole numbers whose sum is 153 is?

sergejj [24]

We can start this problem by finding out what the lowest consecutive number's value is (x). Since consecutive numbers are numbers that are 1 apart from each other, the sum of 9 consecutive numbers would look like

x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) + (x+8)

Since we know that they equal 153,

x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) + (x+8) = 153

Now we combine like terms

9x + 36 = 153

Simplify

9x = 117

x = 13

Now, we need to find what the 5th consecutive number is equal to. The fifth consecutive number is (x+4), so 13 + 4 is 17, meaning that the 5th of 9 consecutive numbers that add up to 153 is 17.

4 0

3 years ago

Read 2 more answers

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,

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