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professor190 [17]
3 years ago
6

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.

You might be interested in
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<em> </em><u><em>Weierstrass substitution</em></u>, 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}

Hence, we showed \sin(x) \text { and } \cos(x)

======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3

\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies  \dfrac{5-5t^2 -24t}{1+t^2}= 3

\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0

t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}

Just note that

\tan\left(\dfrac{x}{2}\right) =  \dfrac{3\pm 8\sqrt{10}}{-2}

and  \tan\left(\dfrac{x}{2}\right) is not defined for x=k\pi , k\in\mathbb{Z}

6 0
3 years ago
Find the distance between the parallel lines y=x+8 and y=x+2
bulgar [2K]

Answer:

Step-by-step explanation:

6

6 0
3 years ago
Which is greater 2/3 or 5/6
marysya [2.9K]
5/6 is greater.
2/3 < 5/6
6 0
3 years ago
I WILL GIVE 100 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT.What is the solution to the inequality 3+5/7x&gt;1+x?
harina [27]

Answer:

C: X <= 7

Step-by-step explanation:

Reimagine the inequality as

3 - 1 >= X - 5/7*X

2 >= 2/7*X

1 >= 1/7 * X

7 >= X

8 0
3 years ago
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