Answer:
25x+ 21y
You multiply x by sweatshirts and y by sweatpants
Step-by-step explanation:
I don't know what sport your talking about or how many people are on the team so im not really sure but hete
1: w=3
2:w=-4
( 1): ''w2'' was replaced by ''w^2''
w^2+w-(12)=0
Answer:
35.09°
Step-by-step explanation:
You can use cos theta to find the value of x.
Let us solve now.
They have already given the lengths of the hypotenuse and adjacent.
Hypotenuse = 33
Adjacent = 27
Let us solve now.
Cos x = Adjacent ÷ Hypotenuse
Cos x = 27 ÷ 33
Cos x = 0.8181
x = Cos⁻¹ 0.8181
x = 35.09°
Hope this helps you :-)
Let me know if you have any other questions :)
Answer: 
This can be written out as g(x) = 1/(x-6) - 4
The "x-6" is in the denominator, while the "-4" is not.
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Explanation:
We'll start with the parent function 
When we shift 4 units down, we're subtracting 4 from the y coordinate of each point on the curve. Which is the same as subtracting 4 from f(x) because y = f(x).
So we have
as an intermediate step.
Then to shift 6 units to the right, we'll replace every x with "x-6". Imagine we kept the h(x) curve completely still, and instead we moved the xy axis 6 units to the left. This would give the illusion of h(x) moving 6 units to the right if we made the xy axis stay still. So that's why we go for "x-6" instead of "x+6".
Therefore, we end up with 
Side note: plugging in x = 6 leads to a division by zero error. This would mean x = 6 is not in the domain.
Answer:
The angle between the given vectors u and v is ![\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7D%5Cright%5D)
Step-by-step explanation:
Given vectors are
and 
Now compute the dot product of u and v:




Now find the magnitude of u and v:









To find the angle between the given vectors

![\theta=cos^{-1}\left[\frac{\overrightarrow{u}.\overrightarrow{v}}{|\overrightarrow{u}|\overrightarrow{v}|}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B%5Coverrightarrow%7Bu%7D.%5Coverrightarrow%7Bv%7D%7D%7B%7C%5Coverrightarrow%7Bu%7D%7C%5Coverrightarrow%7Bv%7D%7C%7D%5Cright%5D)
![=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]](https://tex.z-dn.net/?f=%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B15%7D%7B5%5Ctimes%20%5Csqrt%7B10%7D%7D%5Cright%5D)
![=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]](https://tex.z-dn.net/?f=%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B15%7D%7B5%5Ctimes%20%5Csqrt%7B10%7D%7D%5Cright%5D)
![\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7D%5Cright%5D)
Therefore the angle between the vectors u and v is
![\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7D%5Cright%5D)