When you use a tens fact you round to the nearest ten and 15-8=7. The nearest ten to seven is 10
Answer:10
Answer:
50/96
Step-by-step explanation:
I assume the table is:
![\left[\begin{array}{cc}Result&Total\\1&15\\2&13\\3&16\\4&17\\5&15\\6&20\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DResult%26Total%5C%5C1%2615%5C%5C2%2613%5C%5C3%2616%5C%5C4%2617%5C%5C5%2615%5C%5C6%2620%5Cend%7Barray%7D%5Cright%5D)
Even numbers are 2, 4, and 6. The total number of times she rolled an even number is 13 + 17 + 20 = 50.
So the probability is 50/96.
We are given the following quadratic equation
![f(x)=2x^2+7x-10](https://tex.z-dn.net/?f=f%28x%29%3D2x%5E2%2B7x-10)
The vertex is the maximum/minimum point of the quadratic equation.
The x-coordinate of the vertex is given by
![h=-\frac{b}{2a}](https://tex.z-dn.net/?f=h%3D-%5Cfrac%7Bb%7D%7B2a%7D)
Comparing the given equation with the general form of the quadratic equation, the coefficients are
a = 2
b = 7
c = -10
![h=-\frac{b}{2a}=-\frac{7}{2(2)}=-\frac{7}{4}=-1.75](https://tex.z-dn.net/?f=h%3D-%5Cfrac%7Bb%7D%7B2a%7D%3D-%5Cfrac%7B7%7D%7B2%282%29%7D%3D-%5Cfrac%7B7%7D%7B4%7D%3D-1.75)
The y-coordinate of the vertex is given by
![\begin{gathered} f(x)=2x^2+7x-10 \\ f(-1.75)=2(-1.75)^2+7(-1.75)-10 \\ f(-1.75)=2(3.0625)^{}-12.25-10 \\ f(-1.75)=6.125^{}-12.25-10 \\ f\mleft(-1.75\mright)=-16.13 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20f%28x%29%3D2x%5E2%2B7x-10%20%5C%5C%20f%28-1.75%29%3D2%28-1.75%29%5E2%2B7%28-1.75%29-10%20%5C%5C%20f%28-1.75%29%3D2%283.0625%29%5E%7B%7D-12.25-10%20%5C%5C%20f%28-1.75%29%3D6.125%5E%7B%7D-12.25-10%20%5C%5C%20f%5Cmleft%28-1.75%5Cmright%29%3D-16.13%20%5Cend%7Bgathered%7D)
This means that we have a minimum point.
Therefore, the minimum point of the given quadratic equation is
1) correct
2) -3
3) 15/7
4) -10
Given:
![\sin \theta](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%20%3C0)
![\tan \theta](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3C0)
To find:
The quadrant in which
lie.
Solution:
Quadrant concept:
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II, only
and
are positive.
In Quadrant III, only
and
are positive.
In Quadrant IV, only
and
are positive.
We have,
![\sin \theta](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%20%3C0)
![\tan \theta](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3C0)
Here,
is negative and
is also negative. It is possible, if
lies in the Quadrant IV.
Therefore, the correct option is D.