Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.
Let's see whether they do.
8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.
If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:
8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and
d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.
11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.
Draw a cartesian plane, create a graph with the equation x = y^2 - 2
then substitute numbers into the equation so that it is true, to find points on the graph, e.g. substitute y with 1, you get
x = 1^2 - 2
x = 1 - 2 = -1, so when y = 1, x = -1, this point is (-1, 1)
for the next substitute y with 2,
x = 2^2 - 2
x = 4 - 2 = 2, the point is (2, 2)
you might want to try negative values of y
y = -1, x = (-1)^2 - 2
x = -1 the point is (-1,-1)
then plot the points on the graph
Answer:
Explanation:
The statement is expressed as
y
∝
1
x
inverse means
1
variable
To convert to an equation introduce k, the constant of variation.
y
=
k
×
1
x
⇒
y
=
k
x
To find k use the given condition that y = 8 when x = 4
8
=
k
4
⇒
k
=
4
×
8
=
32
⇒
equation is
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
2
2
y
=
32
x
2
2
∣
∣
∣
−−−−−−−−−−−−
x
=
16
→
y
=
32
16
⇒
y
=
2
when
x
=
16
Answer:
here is the answer... for it
Answer:
Step-by-step explanation:
