Jimmy had five thousand apples and he ate _ he had 97 left how many did he eat

Find two fractions with the difference of 1/5 but with neither denominator equal to 5? I know the answer from Google but how is

Bas_tet [7]

Suppose we try to write out this problem symbolically:

a c 1
-- - --- = --- , b not equal to 5, d not equal to 5
b d 5

Unfortunately, this equation contains 4 unknowns (too many)

We could do a bit of guesswork here:

a c 1
-- - --- = --- , b not equal to 5, d not equal to 5
b d 5

8 6 2
--- - ----- = -----
10 10 10

This is true, but not all that wonderful, because all 3 fractions could be reduced and would then have denominator 5.

6 0

4 years ago

Someone know this please help geometry if you are good at it

polet [3.4K]

Answer:

Option A) outside

$----------$

hope it helps..

have a great day!!!

3 0

3 years ago

A. A frog is climbing out of a well that is 8 feet deep. The frog can climb 4 feet per

Fiesta28 [93]

Answer:

A: 5 hrs. B: 14 hrs. and 40 min.

Step-by-step explanation:

too long to explain.

4 0

4 years ago

Can someone please actually help with with this problem?

Anna11 [10]

Answer:

(a)

$\begin{array}{cccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} & {6} \ \\ {P(x)} & {0} & {4} & {8} & {12} & {16}& {20} & {24} \ \end{array}$

(b)

$P(x) = 4x$

(c)

See attachment for graph

Step-by-step explanation:

Given

P(x) = Perimeter of square

Solving (a): Complete the table

The perimeter of a square is:

$P =4x$

so, the table will be completed by multiplying each value of x by 4.

So, we have:

$\begin{array}{cccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} & {6} \ \\ {P(x)} & {0} & {4} & {8} & {12} & {16}& {20} & {24} \ \end{array}$

Solving (b): The equation

This has already been stated in (a)

$P(x) = 4x$

Solving (c): The graph

See attachment for graph

6 0

3 years ago

A coin is flipped eight times where each flip comes up either heads or tails. How many possible outcomes a) are there in total?

attashe74 [19]

Answer:

There are 256 ways in total.

There are 56 possible outcomes contain exactly three heads.

The possible outcomes contain at least three heads is 219.

The possible outcomes contain the same number of heads and tails are 70.

Step-by-step explanation:

Consider the provided information.

A coin is flipped eight times where each flip comes up either heads or tails.

Part (a) How many possible outcomes are there in total?

Each time we flip a coin it comes up either heads or tail.

Therefore the total number of ways are:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^8=256$

Hence, there are 256 ways in total.

Part (b) contain exactly three heads?

We want exactly 3 heads, therefore,

n=8 and r=3

According to the definition of combination: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$

$\binom{8}{3}=\frac{8!}{3!(5)!}=56$

Hence, there are 56 possible outcomes contain exactly three heads.

Part (c) contain at least three heads?

For 3 heads: $\binom{8}{3}=\frac{8!}{3!(5)!}=56$

For 4 heads: $\binom{8}{4}=\frac{8!}{4!(4)!}=70$

For 5 heads: $\binom{8}{5}=\frac{8!}{5!(3)!}=56$

For 6 heads: $\binom{8}{6}=\frac{8!}{6!(2)!}=28$

For 7 heads: $\binom{8}{7}=\frac{8!}{7!(1)!}=8$

For 8 heads: $\binom{8}{8}=1$

Now add them as shown:

56+70+56+28+8+1=219

Hence, the possible outcomes contain at least three heads is 219.

Part (d) contain the same number of heads and tails?

Same number of heads and tails means that the value of r=4.

Therefore,

$\binom{8}{4}=\frac{8!}{4!(4)!}=70$

Hence, the possible outcomes contain the same number of heads and tails are 70.

6 0

3 years ago