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3 years ago

There are 95 cherry pieces in 5 bags of candy.How many pieces per bag?

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

7 0

The average is (95 pieces / 5 bags) = 19 pieces/bag,
but it may not be consistent in every bag.

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What is the value of q?

malfutka [58]

Triangles QST and RST are similar. Therefore, the following is true:

q s
--- = ---- This results in 10q=rs.
r 10

Also, since RST is a right triangle, 4^2 + s^2 = q^2.
Since QST is also a right triangle, s^2 + 10^2 = r^2.
4 s
Also: ---- = ----- which leads to s^2 = 40
s 10

Because of this, 4^2 + s^2 = q^2 becomes 16 + 40 = 56 = q^2

Then q = sqrt(56) = sqrt(4)*sqrt(14) = 2*sqrt(14) (answer)

5 0

3 years ago

Read 2 more answers

Mohamed decided to track the number of leaves on the tree in his backyard each year The first year there were 500 leaves Each ye

svetlana [45]

Answer:

The required recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

Step-by-step explanation:

Mohamed decided to track the number of leaves on the tree in his backyard each year.

The first year there were 500 leaves

$Year \: 1 = 500$

Each year thereafter the number of leaves was 40% more than the year before so that means

$Year \: 2 = 500(1+0.40) = 500\times 1.4\\$

For the third year the number of leaves increase 40% than the year before so that means

$Year \: 3 = 500\times 1.4(1+0.40) = 500 \times 1.4^{2}\\$

Similarly for fourth year,

$Year \: 4 = 500\times 1.4^{2}(1+0.40) = 500\times 1.4^{3}\\$

So we can clearly see the pattern here

Let f(n) be the number of leaves on the tree in Mohameds back yard in the nth year since he started tracking it then general recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

This is the required recursive formula to find the number of leaves for the nth year.

Bonus:

Lets find out the number of leaves in the 10th year,

$f(10)= 500\times(1.4)^{10-1}\\\\f(10)= 500\times(1.4)^{9}\\\\f(10)= 500\times20.66\\\\f(10)= 10330$

So there will be 10330 leaves in the 10th year.

3 0

3 years ago

Read 2 more answers

What's the product of (1/3x^2y)*(1/6xy^3)

leonid [27]

Answer: $\frac{1}{18x^3y^4}$

5 0

2 years ago

1,000.00 at 4.5% interest for 2 years

siniylev [52]

Each year we add 4.5% interest.
4.5% as a decimal is 0.045.
We multiply by 1.045 to add this interest.
(not 0.045, this would just be the interest and wouldn't keep the whole)

To start: $1,000
Multiply by 1.045 for the first year.
Now we have $1,045.
Multiply by 1.045 again for the second year.
Now we're at 1092.025.
We need to round up to the nearest cent...$1,092.03

4 0

3 years ago

James estimate the product of 48 and 53 is that reasonable

Debora [2.8K]

50 x 50= 2500.............

8 0

3 years ago