7+16y=;1/4 solution or no solution?

Answers

Answer 1

Answer:

Answer:

the answer is, Y= -27/64

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Really easy seventh grade math can I get this problem answered? Thanks!!

Answers

20/8=x/6
x=15
they are similar so the proportion is the same

x = 15

since the figures are similar then the ratios of corresponding sides are equal

$(x)/(6)$ = $(20)/(8)$ ( cross- multiply )

8x = 6 × 20 = 120 ( divide both sides by 8 )

x = 15

The number of milligrams of a certain medicine a veterinarian gives to a dog varies directly with the weight of the dog.If the veterinarian gives a 30-pound dog 5 milligram of the medicine, which equation relates the weight, w, and the
dosage, o?

Answers

Question: The number of milligrams of a certain medicine a veterinarian gives to a dog varies directly with the weight of the dog. If the veterinarian gives a 30-pound dog 3/5 milligram of the medicine, which equation relates the weight,w, and the dosage, d?

Answer: d= 1/50w

Explanation: I took the test in Edgenuity.

Hope this helps!

Write an expression that represents 72 divided by n plus 4

Answers

Answer:

The answer is

$(72)/(n)+4$

Step-by-step explanation:

we know that

A quotient is the division of two numbers

In the expression $72$ divided by n

the numerator is equal to $72$

the denominator is equal to $n$

the quotient is equal to

$(72)/(n)$

The complete expression $72$ divided by n plus $4$ is equal to

$(72)/(n)+4$

72/n+4

Hope that helps

A cable that weighs 6 lb/ft is used to lift 500 lb of coal up a mine shaft 400 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum. (Let x be the distance in feet below the top of the shaft. Enter xi* as xi.)

Answers

The work done to lift the coal is 6.8*10^5 ft-lb

Data;

weight of cable = 6lb/ft
weight of coal = 500lb
length of mine shaft = 400ft

Let the distance (ft) below top of the shaft be represented by x

The weight of the coal to be lifted from mine = 500lb

Work Done

The work done to lift the coal is

$work = force * displacement\nw=fx\nforce f(x) = 500 + 6x\nw_i= f(x_i)=[500+6x_i]\delta x\n$

Taking the summation limit

$w = \lim_(0 \to 400) \sum [500+6x_i]\delta x\n \delta w = f(x) \delta x\n$

we can take the integration of both sides where x will the range from 0 to 400.

$\int \delta w = \int f(x) dx\n\int\limits^(400)_0{(500+6x)} \, dx \nw=[500x+3x^2]_0^4^0^0\nw=500(400-0)+ 3(400^2-0^2)\nw= 680000\nw=6.8*10^5 ft-lb$

From the calculations above, the work done is 6.8*10^5 ft-lb

Learn more on work done on a lift here;

brainly.com/question/10247507

brainly.com/question/7203062

Answer:

8.2 *10^5 ft lb

Step-by-step explanation:

Wcoal = 800*500 = 400000 = 4*10^5 ft lb

$Wrope =\int\limits^(400)_0 {6(400-y)} \, dy$

Using a right Riemann sum

The width of the entireregion to be estimated = 400 - 0 = 400

Considering 8 equal subdivisions, then the width of each rectangular division is 400/8 = 50

F(x) = 6(400-y)

$\left[\begin{array}{cccccccccc}x&0&50&100&150&200&250&300&350 &400\nf(x)&2400&2100&1800&1500&1200&900&600&300&0\end{array}\right]$

Wrope = 50(2100) + 50(1800) + 50(1500) + 50(1200) + 50(900) + 50(600)

+ 50(300) + 50(0) = 420000 = 4.2 * 10^5 ft lb

Note: Riemann sum is an approximation, so may not give a accurate value

work done = Wcoal + Wrope= 4*10^5+ 4.2 * 10^5 = 8.2 *10^5 ft lb

The metal day fair has a recycling center that will a pay 10% early fee on the amount collected plus $2.75 for each metal collected to be recycled .If Isa brings 12.5 kilograms of metal to be recycled which is the closet amount of money she will receive.

Answers

Answer:

$37.8125

Step-by-step explanation:

$2.75 x 12.5 = 34.375

34.375 x 10% = 3.4375

$34.375 + 3.4375 = 37.8125

a radio station claims that the amount of advertising per hour of broadcast time has an average of 17 minutes and a standard deviation equal to 2.7 minutes. You listen to the radio station for 1 hour, at a randomly selected time, and carefully observe that the amount of advertising time is equal to 11 minutes. Calculate the z-score for this amount of advertising time

Answers

Answer:

z-score for 11 minutes of advertising time is $z=(-6)/(2.7)\approx -2.222$

Step-by-step explanation:

Z-scores measure the distance of any data point from the mean in units of standard deviations and are useful because they allow us to compare the relative positions of data values in different samples.

The z-score for any single data value can be found by the formula:

$z=(data \:value- \:mean)/(standard \:deviation)$

From the information given we know:

Data value = 11 minutes
Mean = 17 minutes
Standard deviation = 2.7 minutes

$z=(11-17)/(2.7) = (-6)/(2.7)\approx -2.222$

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