Answer:
a.) Between 0.5 and 3 seconds.
Step-by-step explanation:
So I just went ahead and graphed this quadratic on Desmos so you could have an idea of what this looks like. A negative quadratic, and we're trying to find when the graph's y-values are greater than 26.
If you look at the graph, you can easily see that the quadratic crosses y = 26 at x-values 0.5 and 3. And, you can see that the quadratic's graph is actually above y = 26 between these two values, 0.5 and 3.
Because we know that the quadratic's graph models the projectile's motion, we can conclude that the projectile will also be above 26 feet between 0.5 and 3 seconds.
So, the answer is a.) between 0.5 and 3 seconds.
Answer:
the diagonal of TV equal to 20
Answer: option c
Step-by-step explanation:
Find the x-intercept and y-intercept of each line.
To find the x-intercept, substitute 
into the equation and solve for "x".
To find the y-intercept, substitute 
into the equation and solve for "y".
- For the first equation:
x-intercept
y-intercept
Graph a line that passes through the points (7.25, 0) and (0, 9.66)
- For the second equation:
x-intercept
y-intercept
Graph a line that passes through the points (0.5, 0) and (0, -0.33)
Observe the graph attached. You can see that point of intersection of the lines is (5,3); then this is the solution of the system. Therefore:
Answer:
Option C. is correct
Step-by-step explanation:
In the triangle ABC, vertices are A(12, 8), B(4, 8) and C(4, 14)
AB =
BC =
AC =
In triangle XYZ vertices are X(6,6), Y(4,12) and Z(10, 14)
XY =
YZ =
XZ =
In triangle MNO, vertices are M(4, 16), N(4, 8) and O(-2, 8)
MN =
MO =
NO =
In triangle JKL, vertices are J(14, -2), K(12,2) and L(20, 4)
JK =
KL =
JL =
Now we compare the sides for the congruence
Since AB = MN = 8
BC = NO = 6
AC = OM = 10
So ΔABC ≅ Δ MNO
Option C. is correct
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