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belka [17]
3 years ago
9

Find the value of x.

Mathematics
1 answer:
nekit [7.7K]3 years ago
8 0

Answer:

B

Step-by-step explanation:

Since this right angled triangle has two angles of 45, so according to theorem sides opposite to equal angles are equal

Applying Pythagoras theorem

(4√2)^2=(x)^2+(x)^2

32=x^2+x^2

32=2x^2

x^2=32/2

x^2=16

Taking sq root on both sides we get

x=4

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A sequence of numbers if 0, 12, 24, 36, 48...<br> Drag tiles to complete the statements.
notsponge [240]

Answer:

60, 72, arithmetic

Step-by-step explanation:

48 + 12 = 60

60 + 12 = 72

An 'Arthimetic sequence' is what is presented in this question.

6 0
4 years ago
The 24 students in ms. lees class each collected an average of 18 cans for recycling. then 21 students in mr. galvez's class eac
belka [17]

This can be solve by using the average cans of each student collected and muliply it by the total students. Since for ms. Lee has 24 students and each student collected 18 cans on average, so the total can her class collected on average is 432 cans. For mr galveshas 21 students and collected 25 can per syudents on average, so the total is 525 cans. So 525 – 432 = 93 more cans the class of mr galvez collected

7 0
3 years ago
The elevation of Mt. Everest is 29,028 feet. The elevation of the Dead Sea is –485 feet. What is the difference in the elevation
navik [9.2K]

Answer:

28,543

Step-by-step explanation:

If we take the height of Mt. Everest (29,028) and the height of the Dead sea (-485) and subtract we get the `answer of 28,543.

6 0
4 years ago
Read 2 more answers
How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t
Zarrin [17]

By first principles, the derivative is

\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h

Use the binomial theorem to expand the numerator:

(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n

(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n

where

\dbinom nk=\dfrac{n!}{k!(n-k)!}

The first term is eliminated, and the limit is

\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h

A power of h in every term of the numerator cancels with h in the denominator:

\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)

Finally, each term containing h approaches 0 as h\to0, and the derivative is

y=x^n\implies y'=nx^{n-1}

4 0
3 years ago
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gladu [14]
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