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Help plz!! The function h(t)=-16t^2+22t+4 models the height (h) of a football t seconds after it is thrown. The question ask to

Black_prince [1.1K]

If you're familiar with the quadratic equation
$\frac{ -b \pm \sqrt{b^2-4ac}}{2a}$

well, $\frac{ -b}{2a}$ is the x component of the vertex, in your case, the vertex is the highest point

figure out what t is at the highest point, and then substitute that back into the equation in order to determine height

x=t in this problem

8 0

3 years ago

(-2q^4)(-8q^3b^6)<br>help I need to simplify this problem

Montano1993 [528]

Multiply the constants. Multiply the variables according to the rules of exponents.
$(-2q^{4})(-8q^{3}b^{6}) = ((-2)(-8))q^{(4+3)}b^{6}\\\\=16q^{7}b^6}$

4 0

3 years ago

Which is a solution for 4 x + 8 = 3 x + 14?

Liula [17]

Answer:

x=6

Step-by-step explanation:

Step 1:

Subtract 3x from both sides:

4x + 8 - 3x = 3x + 14 - 3x

x + 8 = 14

Step 2:

Subtract 8 from both sides:

x + 8 - 8 = 14 - 8

x = 6

4 0

3 years ago

Read 2 more answers

Pls help! evaluate the sine of the real number, pls give exact answers

solong [7]

Answer:

$-\frac{\sqrt2}{2} \\$ or exact is −0.70710678...

Step-by-step explanation:

Getting the point!

If we look at the unit circle and we find $-\frac{3\pi }{4}$ we see it is in the 3rd quadrant.
We know that all points in the 3rd quadrant have both negative x and negative y values.
We know that all radians with the base of 4 on the unit circle have the points of $(\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{2})$
Because we are in the 3rd quadrant and both x and y are negative we know the point of $-\frac{3\pi }{4}$ is $(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$