When the wheels of a bicycle rotate 30 times the bicycle moves a distance of 175 feet what is the rate

PLEASE SOMEONE HELP ILL MARK BRAINLIEST AND 5 STARS 2.1/x+ 1 4/5= 3/10x

enot [183]

Answer:

the answer is .3 or 3/10

Step-by-step explanation:

8 0

3 years ago

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A baker buys 19 apples of two different varieties to make pies. The total cost of the apples is $5.10. Granny Smith apples cost

Natasha2012 [34]

The baker bought 7 gala apples, and 12 granny smiths apples

Solution:

Let "x" be the number of gala apples bought

Let "y" be the number of granny smith apples bought

Cost of 1 gala apple = $ 0.30

Cost of 1 granny smith apple = $ 0.25

A baker buys 19 apples of two different varieties to make pies

Therefore,

number of gala apples bought + number of granny smith apples bought = 19

x + y = 19 --------- eqn 1

The total cost of the apples is $5.10

Therefore, we can frame a equation as:

number of gala apples bought x Cost of 1 gala apple + number of granny smith apples bought x Cost of 1 granny smith apple = 5.10

$x \times 0.30 + y \times 0.25 = 5.10$

0.3x + 0.25y = 5.1 -------- eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

x = 19 - y ---------- eqn 3

Substitute eqn 3 in eqn 2

0.3(19 - y) + 0.25y = 5.1

5.7 - 0.3y + 0.25y = 5.1

5.7 - 0.05y = 5.1

0.05y = 5.7 - 5.1

0.05y = 0.6

y = 12

Substitute y = 12 in eqn 3

x = 19 - 12

x = 7

Thus the baker bought 7 gala apples, and 12 granny smiths apples

6 0

3 years ago

Someone please help me

xxTIMURxx [149]

I think its 6 i'm not sure hope this might help

7 0

3 years ago

a gym charges a one time fee of $50 and monthly membership charge of $25 the total cost c of being a member of the gym is given

marissa [1.9K]

A. The slope is 25 and it is the rate of change per month. You will pay this additional amount if you go to the gym every month.
b. The y-intercept is 50. This amount is the constant and does not change since it is a one time fee for the year.

4 0

4 years ago

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Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

PolarNik [594]

Answer:

Using backward reasoning we want to show that "We can never get nine 0's".

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

There were 9 0's:

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

There were 9 1's:

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

5 0

3 years ago