Simplify -3a(5a^2a+3)-5a

Consider the first four terms of the sequence below. -3, -12, -48, -192, . . . What is the 8th term of this sequence? A. -49,152

kolbaska11 [484]

Answer:

A

Step-by-step explanation:

there is a common ratio between consecutive terms , that is

- 12 ÷ - 3 = - 48 ÷ - 12 = - 192 ÷ - 48 = 4

this indicates the sequence is geometric with nth term

$a_{n}$ = a₁ $r^{n-1}$

where a₁ is the first term and r the common ratio

here a₁ = - 3 and r = 4 , then

a₈ = - 3 × $4^{7}$ = - 3 × 16,384 = - 49,152

7 0

2 years ago

Round up 0.12 to the nearest tenth

nata0808 [166]

0.12 rounded to the nearest tenth is 0.10

8 0

3 years ago

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

so

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

so

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

3 years ago

A student thinks that (1, 2) is a solution to y = 2x - 3. While checking the solution, the student gets 1 = 2(2) -3. What error

BabaBlast [244]

He put in the value of x into y
and the value of y in x
if that confuse you here this is what it will look like
(x,y)->(y,x)

6 0

3 years ago

A local carpet company has been hired to carpet a planetarium which is in the shape of a circle. If the radius of the planetariu

Vika [28.1K]

Pie r squared is the area of a circle, so 9^2(3.14etc) = so 255 square yards. So 255 x 15 = $3,825

8 0

3 years ago