Answer:
1/sqrt10
Step-by-step explanation:
1) Find out cosA using formula (cosA)^2+(sinA)^2=1
The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5
So cosA=-4/5 or cosA=4/5.
Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.
2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.
(sinA/2)^2= 0.1
sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10
But 270°< A< 360°, then 270/2°<A/2<360/2°
135°<A/2<180°, so sinA/2 must be positive and the only correct answer is
sin A/2= 1/sqrt10
Answer:
m=-4
y-intercept=8
x-intercept=2
Step-by-step explanation:
Please see attachment.
y-intercept=8
The y-intercept is where the line crosses the y-axis.
x-intercept=2
The x-intercept is where the line crosses the x-axis.
m=-4
If you do not know how to find slope by looking at graph, use slope formula.
(y2-y1)/(x2-x1)
Choose 2 points. You may use the x and y intercepts.
(2,0) and (0,8)
(8-0)/(0-2)=8/-2=-4
Hope this helps!
Answer:
I think the answer for this is the Dimm slots
9514 1404 393
Answer:
a. It represents both a relation and a function
Step-by-step explanation:
The expression on the right is a polynomial of degree 2.
Any polynomial in a single variable is a function.
Any function is a relation.
So, ...
it represents both a relation and a function
When you plot the offered points on the graph, you find only one of them falls inside the shaded region:
... D. 43 pepperoni slices and 25 cheese slices