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Let's solve your equation step-by-step.
(
4
x
−
3
)
(
(
x
−
2
)
2
)
=
4
x
3
−
19
x
2
+
4
x
−
12
4
x
3
−
19
x
2
+
28
x
−
12
=
4
x
3
−
19
x
2
+
4
x
−
12
Step 1: Subtract 4x^3 from both sides.
4
x
3
−
19
x
2
+
28
x
−
12
−
4
x
3
=
4
x
3
−
19
x
2
+
4
x
−
12
−
4
x
3
−
19
x
2
+
28
x
−
12
=
−
19
x
2
+
4
x
12
Step 2: Add 19x^2 to both sides.
−
19
x
2
+
28
x
−
12
+
19
x
2
=
−
19
x
2
+
4
x
−
12
+
19
x
2
28
x
−
12
=
4
x
−
12
Step 3: Subtract 4x from both sides.
28
x
−
12
−
4
x
=
4
x
−
12
−
4
x
24
x
−
12
=
−
12
Step 4: Add 12 to both sides.
24
x
−
12
+
12
=
−
12
+
12
24
x
=
0
Step 5: Divide both sides by 24.
24
x
24
=
0
24
x
=
0
Answer:
x
=
0
Answer:
9 mile/ 1 hour
Step-by-step explanation:
Because the squirrel ran 9 miles in 1 hour, so u have to divide 9 miles by 1 hour.
Answer:
The slope is 10/5
Step-by-step explanation:
The slope is the rise over run, or how many it goes up/down over how many it goes left/right.
If you keep adding to -2 all the way until you get 8, you'd get 10, since you'd add 2 to get to 0 and 8 to get to 8
If you keep adding to -5 all the way to 0, you'd get 5.
Hope this helps!
Answer:
a. y(t) = - 1/2gt² + y₀ b.
Step-by-step explanation:
Let the quadratic function y(t) = v₀t - 1/2gt² + y₀ represent the quadratic function that models the height above the ground of the projectile.
a. Maximum Height
At maximum height, the velocity, v₀ = 0, so substituting v₀ = 0 into the equation, we have
y(t) = v₀t - 1/2gt² + y₀
y(t) = 0 × t - 1/2gt² + y₀
y(t) = 0 - 1/2gt² + y₀
y(t) = - 1/2gt² + y₀
b. Time when the projectile reaches the ground
The time when the projectile reaches the ground is gotten when y(t) = 0, So
y(t) = v₀t - 1/2gt² + y₀
0 = v₀t - 1/2gt² + y₀
re-arranging, we have
1/2gt² - v₀t - y₀ = 0
Using the quadratic formula,