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3 years ago

What ratio is equivalent to 1/7

2 answers:

4 0

Any ratio which can be simplified to 1/7 is equivalent to it. So, for example, if you multiply this fraction by 2, you get 2/14, which is equivalent. Or 3/21, or 4/28, or 5/35...

Valentin [98]3 years ago

3 0

Any ratio that is of the form
1*k 1 k 1
---- = ------------ * ------- =---------
7*k 7 k 7

as k/k =1
so if you find some ratio give to you such that u can factor our some number k from both numerator and denominator and have 1/7 left out, then it is equivalent... for example

2 1*2 1 2 1
--- =-----=----- * ------ =-----
14 7*2 7 2 7

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Find the future value of $2000 invested at annual rate of 2.4% compounded quarterly for 8 years. Round your answer to hundredths

MatroZZZ [7]

Answer: $2421.95

Step-by-step explanation:

A = p(1 + r/n) ^nt

A= 2000(1+0.024/4)^4x8

A= 2421.95

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4 years ago

Is anyone able to help me with this question:

Serga [27]

Answer:

not sure but this info might help you.

Step-by-step explanation:

the diagonal in the rectangle splits it into two right angled triangles, hence the hypotenuse is the diagonal.

once you find the measurement, we all know that a square has all sides equal and the diagonal splits it into 2 equilateral triangles. so, the side of the square =the diagonal's measurement.

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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

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

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

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

$0 \leq x \leq 48.6\ ft$

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

3 years ago

Help quick Given f(x) and g(x) = k⋅f(x), use the graph to determine the value of k.

Salsk061 [2.6K]

Answer:

k = -3

Step-by-step explanation:

F(x) has a positive slope

g(x) has a negative slope

Let x = -3

The y value goes from 1 to -3

g(x) = k⋅f(x)

-3 = k(1)

-3 = k

6 0

3 years ago

Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)

ehidna [41]

This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ], slope is 86.3

111) for [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope = ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11) [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111) [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

3 0

3 years ago