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3 years ago

Why would we start at (0,200)?

Mathematics

1 answer:

Murrr4er [49]3 years ago

3 0

Answer:

Because 200 was the initial amount. He was not growing $0 by 10% every because that would mean he started with nothing and still has nothing. So we start with 200 because that is what he started with and is growing.

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If the 6th term of an A.P. is 46, find the sum of first 11 terms.

zaharov [31]

Answer:

506

Step-by-step explanation:

:. t6 = 46 ... [ Given ]

tn = a+(n-1) d ... [ Formula ]

:. t6 = a+(6-1) d

:. 46 = a+5d ... [1]

Sn = n/2 [2a+(n-1)d] ... [ Formula ]

:. S11 = 11/2 [2a+(11-1) d]

:. S11 = 11/2 [2a+10d]

:. S11 = 11/2x2 [a+5d]

:. S11 = 11[46] ... [ From (1)]

:. S11 = 506

5 0

3 years ago

On level ground, a vertical rod 12 feet tall casts a shadow 4 feet long, and at the same time a nearby vertical flagpole casts a

Savatey [412]

$\frac{Actual}{Shadow}$ = $\frac{Actual}{Shadow}$
$\frac{12}{4}$ = $\frac{Actual₂}{12}$
Actual = 36 feet

7 0

3 years ago

Read 2 more answers

using the numbers 5, 8, and 24, create a problem using no more than four operations (addition, subtraction, division, square, sq

irga5000 [103]

24^2 / (8+5)

576 / 13

= 44.3

7 0

3 years ago

Please help!! Very urgent..thank you!!

erastova [34]

Answer : 4 times

Here it's given that ,

The height and base of the butterfly sitting on the stem (red butterfly) is two times greater than the height and base of the butterfly sitting on the flower .

And we need to find out how many times the area of red winged butterfly is greater than that of sitting on the flower (blue butterfly) .

Let us take ,

base of blue butterfly be b
height of blue butterfly be h
Area be A .

Then ,

base of red butterfly will be 2b .
height of red butterfly will be 2h .
Area be A' .

We know that ,

→ area of the triangle = 1/2 × base × height

So that ,

→ A/A' = (1/2 * b * h) ÷ (1/2 *2b *2h)

→ A/A' = bh/4bh

→ A/A' = 1/4

→ A' = 4A

Henceforth the area of blue butterfly is 4 times greater than that of red winged butterfly .

I hope this helps.

6 0

2 years ago

Oliver works at a bookstore.He packed 20 identical paperbacks and 9 identical textbooks in a box. The total mass of the books wa

olchik [2.2K]

If we let p and t be the masses of the paper and textbook, respectively, the equations that would best represent the given in this item are:
(1) 20p + 9t = 44.4
(2) (20 + 5)p + (9 + 1)t = 51

The values of p and t from the equation are 0.6 and 3.6, respectively. Thus, each paperback weighs 0.6 pounds and each textbook weighs 3.6 pounds.

7 0

3 years ago