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3 years ago

15

Simplify 3 square root of 5 end root minus 2 square root of 7 end root plus square root of 45 end root minus square root of 28.

(1 point)

2 answers:

KonstantinChe [14]3 years ago

7 0

3root5 - 2root7 + root45 - root28

2 root 7 is the same as root28.
3root5 is the same as root45.

2root7 is the answer

pshichka [43]3 years ago

5 0

Answer:

6√5 - 4√7

Step-by-step explanation:

Edge 2020. 100%

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Which of the following statements is true?

Arada [10]

Answer:

its c i tink but im not completly sure

Step-by-step explanation:

5 0

2 years ago

AJKL and ALMN are shown.

seropon [69]

Answer:

mLNM=63

Step-by-step explanation:

triangle JKL is isocelese so mKJL and KLJ are the same

180-72=108

108/2=54

so both mKLJ and mMLN are equal

and since triangle LMN is an isoceles then mLNM and mLMN are equal to each other

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7 0

2 years ago

P=2^(43112609)-1 Suppose you have traveled to a faraway place where the base of the number system is 37 . (The inhabitants of th

AfilCa [17]

Answer:

8,275,842 digits

Step-by-step explanation:

One more than the floor of the logarithm to the base 37 gives the number of digits. The change of base formula can be used.

$\log_{37}{(2^{43112609}-1)}\approx\dfrac{43112609\log{2}}{\log{37}}\approx 8275841.2$

Expressed in the base-37 number system, P has 8275842 digits.

6 0

3 years ago

An instructor who taught two sections of engineering probability last term, the first with 20 students and thesecond with 30, de

Oliga [24]

Answer:

0.207

Step-by-step explanation:

This is an hypergeometric distribution problem

An hypergeometric distribution has the same sense as the discrete probabilities of binomial distribution, but unlike binomial distribution, hypergeometric distribution does not allow replacement.

Binomial distribution expresses the probability of picking k objects from n with replacement, but hypergeometric distribution expresses picking k objects from n without replacement, with the finite total population, N, containing K objects.

It is expressed mathematically as

h(k: n, K, N) = (ᴷCₖ)(ᴺ⁻ᴷCₙ₋ₖ)/(ᴺCₙ)

where

k = number of students in the 2nd section required to be in the first 15 graded projects (number of successes) = 10

n = total number of first graded projects (number of trials) = 15

K = number of students in the 2nd section of the class = 30

N = total number of students = 50

h(10: 15, 30, 50) = (³⁰C₁₀)(⁵⁰⁻³⁰C₁₅₋₁₀)/(⁵⁰C₁₅)

h(10: 15, 30, 50) = (³⁰C₁₀)(²⁰C₅)/(⁵⁰C₁₅)

= (30,045,015)(15,504)/(2,250,829,575,120)

P(X = 10) = 0.207

Hope this Helps!!!

7 0

2 years ago

Find the perimeter of rectangle whose length is 40 and the diagonal is 41 cm<br>

Bond [772]

Answer:

<h2> P = 98 cm</h2>

Step-by-step explanation:

Given one side and diagonal of rectangle we can use Pythagorean theorem to calculate the other side of it.

$L=40\, cm\\D=41\,cm\\W=\ ?\\\\W^2+L^2=D^2\\\\W^2+40^2=41^2\\\\W^2+1600=1681\\\\W^2=81\\\\W=9$

Perimeter:

P = 2W + 2L = 2•40 + 2•9 = 80 + 18 = 98

4 0

3 years ago