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3 years ago

(16 points please help)

Mathematics

1 answer:

zaharov [31]3 years ago

7 0

C. $1,120.91 To get this, just subtract from the beginning balance the withdrawals and add the deposits.

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Cual es la diferencia entre crecimiento lineal y exponencial.ejemplos

Sedbober [7]

Answer:

English please

Step-by-step explanation:

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3 years ago

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Olin [163]

Answer:

Step-by-step explanation:

if A = 1 2 b then it comes to h

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3 years ago

10) A rhombus has a diagonal 8.6 mm long and an area of 81.7 mm. what is the perimeter?

BaLLatris [955]

Answer:

P=41.72

Step-by-step explanation:

S=ACxDB/2

81.7=8.6xDB/2

81.7=4.3xDB|:4.3

19(mm)=DB

DO=19/2=9.5

OC=8.6/2=4.3

(O is the center of the rhombus, where two diagonals meet)

a²+b²=c² (DO²+OC²=DC²)

9.5²+4.3²=c²

90.25+18,49=c²

√108,74=√c²

c≈10.43

P=4c

P=4x10.43

P=41.72

Hope it helps:)

6 0

2 years ago

Multi-Step Equation 2: Solve for x. Show your steps. 6 - 5x + 2 = 3x - 12 + 2x

amm1812

Answer:

The answer is x = 8/5

Step-by-step explanation:

6 - 5x + 2 = 3x - 12 + 2x

4 - 5x = 5x - 12

4 = 10x - 12

16 = 10x

8/5 = x

4 0

2 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

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

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

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

$0.028x=0.68$

$x=24.3$

Alternative method

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

$0 \leq x \leq 48.6\ ft$

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

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

2 years ago