We have that
<span>3x-2y=8 -----> equation 1
2x+3y=Q----> equation 2
the solution is the point </span><span>(4,2)
in the equation 2 substitute the value of
x=4
y=2
so
</span>2x+3y=Q------> 2*4+3*2=Q-------> Q=8+6------> Q=14
<span>
the answer is
Q=14
</span>
Answer:
16
Step-by-step explanation:
Well if you think about it, if she has a total of 10 attempts she has to make 8 of them to make it on the team. Now the total attempts is double so we double the number of throws she needs to make in order to make the team so to do this we would do 8*2 which is 16. She needs to make 16 throws to make it on the team.
Consider the function

, which has derivative

.
The linear approximation of

for some value

within a neighborhood of

is given by

Let

. Then

can be estimated to be

![\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B63.97%7D%5Capprox4-%5Cdfrac%7B0.03%7D%7B48%7D%3D3.999375)
Since

for

, it follows that

must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function

. This means the estimated value is an overestimation.
Indeed, the actual value is closer to the number 3.999374902...
Answer: So if its a sphere the answer its 7238.2.
If its a cylinder the answer is 371.7.
If its a triangular shape the answer is
123.9.
Subtract the known trinomial from the total sum:
6x2 - 5x + 4 - 4x2 + 3x - 2 = 2X^2 -8x +6
I hope this helps :)