Ask question

Ask question

All categories

mihalych1998 [28]

2 years ago

10

An apple is launched directly upward at 64 feet per second from a platform 80 feet high. The equation for this apples height h a

nd time t seconds after launch is h=-16t^2+64t+80. Find the maximum height of the apple

1 answer:

Ronch [10]2 years ago

3 0

Answer:

144

Step-by-step explanation:

X=(-B/2A)

X=-64/2(-16)

X=2

H=-16(2)^2+64(2)+80

H=144

You might be interested in

What is the solution to the inequality shown below?

lesantik [10]

To solve the inequality, you need to isolate/get the variable "p" by itself in the inequality:

5p + 26 < 72 Subtract 26 on both sides

5p + 26 - 26 < 72 - 26

5p < 46 Divide 5 on both sides to get "p" by itself

$\frac{5p}{5}$

$p$

p < 9.2 Your answer is A

8 0

2 years ago

Consider the equation x2+4x+9=0 in standard form. Which equation shows the coefficients a, b, and c correctly substituted into t

Afina-wow [57]

Answer:

<h2>x = -2+i√5 and -2i-√5</h2>

Step-by-step explanation:

The general form of a quadratic equation is ax²+bx+c = 0

Given the quadratic equation x²+4x+9=0 in its standard form, on comparing with the general equation we can get the value of the constant a, b and c as shown;

ax² = x²

a = 1

bx = 4x

b = 4

c = 9

The quadratic formula is given as x = -b±√(b²-4ac)/2a

Substituting the constant;

x = -4±√(4²-4(1)(9))/2(1)

x = -4 ±√(16-36)/2

x = -4±√-20/2

x = -4±(√-1*√20)/2

Note that √-1 = i

x = -4±(i√4*5)/2

x = (-4±i2√5)/2

x = -4/2±i2√5/2

x = -2±i√5

The solution to the quadratic equation are -2+i√5 and -2i-√5

5 0

3 years ago

In circle O, central angle AOB measures

GalinKa [24]

Answer: 6 (pi)

Step-by-step explanation:

Feta/3 = (arc)/Radius

5 0

2 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

2 years ago

Help please 10 points!

Alinara [238K]

Answer:

I'm not sure, but it might be 120.

4 0

2 years ago