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3 years ago

Determine which is the better buy

Mathematics

1 answer:

den301095 [7]3 years ago

6 0

Malay ko

Icompute mo para alam mo

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Bret has a fish tank that holds 320 fluid ounces of water. On average, the water evaporates from the tank at a rate of 3.84 ounc

Lana71 [14]

320 - (3.84 x 14) = 266.24 oz

He was gone for two weeks means he was gone for 14 days. If it drains 3.84 per day, he will lose oz of water equivalent to (3.84 x 14), resulting in him having 266.24 oz when he gets back.

8 0

3 years ago

Read 2 more answers

It takes 40 min for a bus to cover 36 miles. A car travels 1.4 times faster. When will they meet if they started 36 miles apart

sveticcg [70]

Bus:

40 min = 36 miles

1 min = 36 ÷ 40 = 0.9 mile

Car:

1 min = 0.9 x 1.4 = 1.26 miles

Total distance covered by both vehicles in 1 min:

1.26 + 0.9 = 2.16 miles

Time needed for both of them to cover 36km:

36 ÷ 2.16 = 16 2/3 min

Answer: They will meet 16 2/3 mins after they have started.

5 0

3 years ago

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The distance on a map between two locations is

Natalka [10]

Answer:

54 miles

Step-by-step explanation:

lets work in steps here:

1 in (24 miles) + 1 in (24 miles) = 2 in OR 48 miles

PLUS

**0.25 in (6 miles)

= 48+6 miles = 54 miles

** you know that 0.25=6 miles because 24/6 is 4. 100/4 is 0.25. simple division.

5 0

3 years ago

Read 2 more answers

I need help with this question thanks :) the options are 45 33.75 50 not here

Degger [83]

Answer:

UV = 45

Step-by-step explanation:

ZY is the perpendicular bisector of UV , then

UV = 2 × UY = 2 × 22.5 = 45

6 0

3 years ago

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago