<img src="https://tex.z-dn.net/?f=%282-%5Csqrt%7B3%7D%20%29%2A%282%2B%5Csqrt%7B%7D%203%29%3D1" id="TexFormula1" title="(2-\sqrt{

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alukav5142 [94]

3 years ago

3} )*(2+\sqrt{} 3)=1" alt="(2-\sqrt{3} )*(2+\sqrt{} 3)=1" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

goblinko [34]3 years ago

8 0

Here is your answer

fomenos3 years ago

4 0

This Equation would be True. They do equal 1.

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1. Solve using long division.

Ann [662]

Answer:

9x^2 - 3x - 2, R -7.

Step-by-step explanation:

x - 5) 9x^3 - 48x^2 + 13x + 3(9x^2 - 3x - 2 <---- Quotient.

9x^3 - 45x^2

- 3x^2 + 13x

-3x^2 + 15x

-2x + 3

-2x + 10

-7 <----- Remainder

5 0

3 years ago

Neal thought of a number. When he divided 47 by his number the remainder was 5. What can Neal's number be? Find all possible ans

Alexxandr [17]

47x5= 235
235 divided by 5 =47
Neals number could be 47

6 0

3 years ago

Read 2 more answers

6. Using Table 11-2 from your text, calculate the present value (principal) and the compound interest given: Compound Amount = $

storchak [24]

Answer:

The Principal is $877.19 And Compound Interest is $372.81

Step-by-step explanation:

Given as :

Amount after investment = $1250

The time period of investment = 6 years

The nominal rate = 6% compounded semiannually

Let the principal = P

From compounded method

Amount = Principal × $(1 + \frac{Rate}{2\times 100})^{2\times Time}$

Or, $1250 = P × $(1 + \frac{6}{2\times 100})^{2\times 6}$

Or, $1250 = P × $(1.03)^{12}$

Or, $1250 = P × 1.425

∴ P = $\frac{1250}{1.425}$

I.e P = $877.19

So , Principal = $877.19

Now, Compound Interest = Amount - Principal

= $1250 - $877.19

Or, CI = $372.81

Hence The Principal is $877.19 And Compound Interest is $372.81 Answer

8 0

3 years ago

A pro basketball player is a poorâ free-throw shooter. Consider situations in which he shoots a pair of free throws. The probabi

zlopas [31]

Answer:

The probability that he makes one of the two free throws is 0.38

Step-by-step explanation:

Hello!

Considering the situation:

A pro basketball player shoots two free throws.

The following events are determined:

A: "He makes the first free throw"

Ac: "He doesn't make the first free throw"

B: "He makes the second free throw"

Bc: "He doesn't make the second free throw"

It is known that

P(A)= 0.48

P(B/A)= 0.62

P(B/Ac)= 0.38

You need to calculate the probability that he makes one of the two free throws.

There are two possibilities, that "he makes the first throw but fails the second" or that "he fails the first throw and makes the second"

Symbolically:

P(A∩Bc) + P(Ac∩B)

P(A)= 0.48

P(Ac)= 1 - P(A)= 1 - 0.48= 0.52

P(Ac∩B) = P(Ac) * P(B/Ac)= 0.52*0.38= 0.1976≅ 0.20

P(A∩B)= P(A)*P(B/A)= 0.48*0.62= 0.2976≅ 0.30

P(A)= P(A∩B) + P(A∩Bc)

P(A∩Bc)= P(A) - P(A∩B)= 0.48 - 0.30= 0.18

Step 3

P(Ac∩B) + P(A∩Bc) = 0.20 + 0.18= 0.38

I hope this helps!

8 0

3 years ago

Write the integer as a product of two integers such that one of the factors is a perfect square. 27

butalik [34]

Hope this helps....

The factors of 27 are all #'s btwn -27 and 27 that / 27 evenly.

-27, -9, -3, -1, 1, 3, 9, 27

3 0

4 years ago