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3 years ago

How do u solve this ?

Mathematics

1 answer:

Nonamiya [84]3 years ago

6 0

You add all the numbers that you get from multiplying all the numbers

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the rectangular grassy plot 126 m by 48m has a gravel path 1.5m wide a road 2.5m wide is constructed all around it on inside of

dusya [7]

Answer:

= Sh. 2281.50

Step-by-step explanation:

Area of the path;

(126 × 2.5 ) + (2.5 × (48 - (2.5×2)

= 315 + (2.5 × 43)

= 315 + 107.5

= 422.5 m²

But; the cost of graveling is sh 5.40 per square meter

Therefore; The cost of graveling the whole path;

= 422.5 × 5.40

<u>= Sh. 2281.50</u>

6 0

4 years ago

Please help!!! ill give brainliest

yawa3891 [41]

Answer:

Step-by-step explanation:

divide the number then add on

7 0

3 years ago

Read 2 more answers

Can you help me out it would be much appreciate thank you

brilliants [131]

Answer:

The answer to the question is 50.28 feet²

Step-by-step explanation:

Here, is why the answer to the problem is 13.72!

The area of the shaded region is:-

Area of total region - Area of the circular region

Area of total region is 8² so it is 64 feet²

Area of Circular region formula is πr²

Substitute the values, for the formula and it is π4² = 22/7 × 16

→ 352/7

⇒ 13.72 Feet²

Therefore, the area of the shaded region is 13.72 Feet²

(Now, we subtract the two values)

64 - 13.72

⇒ 50.28 Feet²

I hope this answer helps!

7 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

4 years ago

In 1875 an estimated twelve-and-a-half trillion locusts appeared in Nebraska.Write the number of locusts in scientific notation.

oksano4ka [1.4K]

Answer:

12.5 x 10^12

Step-by-step explanation:

12,500,000,000,000

12 zeroes following 12.5

3 0

4 years ago