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shusha [124]
3 years ago
11

William’s typical running speed is 6 mi/h. The difference between his fastest speed and his typical speed is 1.5 mi/h. The equat

ion f minus 1.5 = 6 can be used to represent this problem, where f is William’s fastest speed. What is William’s fastest speed?
Mathematics
2 answers:
mrs_skeptik [129]3 years ago
6 0

Answer:

5.5

Step-by-step explanation:

Gnom [1K]3 years ago
3 0

Answer:

7.5

Step-by-step explanation:

i finished the test:)

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You have 15 coins in your pocket that are quarters or nickels. They total $2.75. How many of each coin do you have?
azamat

Answer: There are 5 nickels and 10 quarters.

Step-by-step explanation:

Let x = Nickels , y= Quarters.

Then x+y = 15

⇒ y=15-x  (i)

1 nickel =  5 cents , 1 quarter = 25 cents

$2.75 = 275  cents    [ 1 dollar = 100 cents ]

Then, 5x+25y= 275    (ii)

Substitute value of y from (i)

5x+25 (15-x)=275

⇒ 5x + 375 - 25x =275

⇒ -20x= 275-375

⇒ -20x= -100

⇒  x = 5   [divide both sides by -20]

Put this in (i) , we get y= 15-5=10

Hence, there are 5 nickels and 10 quarters.

4 0
3 years ago
PLS HELP ASAP
djverab [1.8K]
Substitute the value 350 in for your velocity:
350 = 20 * sq rt(273+t)
divide both by 20 --> 17.5 = sq rt(273 +t)
square both to cancel out the sq rt --> 306.25 = 273 +t
then just subtract 273 from both to get 33.25 = t
3 0
3 years ago
Convert 25/9 into a decimal
Serga [27]
2.778 will be the answer

8 0
4 years ago
Read 2 more answers
What are the possible values of x in x2 + 3x + 3 = 0?
stealth61 [152]
Hello,

x^2+3x+3=0\\

\Delta=9-4*3=-3\\

x= \dfrac{-3-i*\sqrt{3}}{2} \ or\ x= \dfrac{-3+i*\sqrt{3}}{2} \\


3 0
3 years ago
Read 2 more answers
Please explain how to do #6 i know the work is right but i don’t understand it
mariarad [96]
Since this is part of a packet, there's a lot of prior knowledge you need to understand the question.
1) The angle of elevation is the angle between the horizontal and the light of sight from the object to the observer (see picture 1).

2) A lot of the units are from work done by an Indian astronomer. The line of sight (light blue in picture 1) is r = 3438 (I think you might have forgotten to finish writing the number). r, or 3438, is the distance from the earth to the sun used by Indian Astronomers. 

3) The table shows units in jya(θ<span>°). Jya is a Sanskrit (ancient Indian language) word. It stands for the length of half of a chord that connects two endpoints of a circle (kinda confusing and more information than you need to know - but it's the green line in picture 2).
What's important is: </span>jya(θ°) = r*sin(θ°)



Now let's tackle the problem:
1) You know that the angle of elevation is 67.5°. Using the chart, the only important value is the second from the bottom. jya(67.5°) = 3177. 

2) Remember that jya(θ°) = r*sin(θ°). 
Also remember that sine of an angle is \frac{opposite}{hypotenuse}. In this problem, r, the distance from the earth to the sun, happens to be the hypotenuse of the triangle since its the longest edge. 

3) Notice that you're dividing jya(θ°) (aka r*sin(θ°)) by r = 3438. That will give you sin(θ°) by itself. Then you're multiplying that sin(θ°) by 1 AU, which we are told is the actual distance of the hypotenuse/distance from earth to sun. Since sine = \frac{opposite}{hypotenuse} that gives you the apparent length of the opposite side = apparent height of the sun.

The math is a bit weird, but you're basically multiplying sine by the apparent distance 1AU to get the apparent height, instead of multiplying sine by the estimated height r = 3438. Dividing 3177 by 3438 gives you sin(67.5°). Then multiplying sin(67.5°) by the apparent hypotenuse, 1AU, gives you the apparent height. That's how I'm understanding it!

6 0
3 years ago
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