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3 years ago

Is 4/7 a real number

2 answers:

7 0

Yes it is a real number. All numbers including 0 are real except for roots of negative numbers. For example, the square root of 5 would be real but the the square root of -5 would not.

Margarita [4]3 years ago

3 0

Answer:

Yes, it is a real number.

Step-by-step explanation:

All numbers are real numbers.

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ABCD is a quadrilateral inscribed in a circle, as shown below:

malfutka [58]

Answer:

Option 4: <u>(2x + 16)° + (x)° = 180°</u>

Step-by-step explanation:

See the attached figure.

For a quadrilateral inscribed in a circle

The opposite angles are supplementary.

So, ∠D + ∠B = 180°

From the figure:

∠B = x° and ∠D = (2x+16)°

∴ <u>(2x + 16)° + (x)° = 180°</u>

4 0

3 years ago

(6+√25)/(4-√3) can be written in the form of (2+s√3)/13 where r and s are both integers what are the values of r and s

cestrela7 [59]

Answer:

r = 33

s = 18

Step-by-step explanation:

Given expression is $\frac{6+\sqrt{27}}{4-\sqrt{3}}$ .

Multiply numerator and denominator with the conjugate of (4 - √3).

$\frac{(6+\sqrt{27})(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}$

= $\frac{4(6+\sqrt{27})+\sqrt{3}(6+\sqrt{27})}{4^{2}-(\sqrt{3})^2}$

= $\frac{24+4\sqrt{27}+6\sqrt{3}+\sqrt{81})}{16-3}$

= $\frac{24+12\sqrt{3}+6\sqrt{3}+\sqrt{81})}{16-3}$

= $\frac{24+18\sqrt{3}+9}{13}$

= $\frac{33+18\sqrt{3}}{13}$

Now compare this expression with $\frac{r+s\sqrt{3} }{13}$

r = 33

s = 18

3 0

3 years ago

What is the total amount that Matthew's bank will receive after lending him $8,000 for four years at an interest rate of 6 perce

xxMikexx [17]

480/year. That's the correct math

5 0

4 years ago

Elephant eats 250 pounds of food every day how many days would it take an elephant to eat a ton of food

ankoles [38]

1 ton= 2,000 pounds

250 x 8 = 2,000

final answer: it would take the elephant 8 days to eat a ton of food

*HOPE I HELPED!*

5 0

3 years ago

What is <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bx%2B2%7D%20%2B%5Cfrac%7B6%7D%7Bx-5%7D%20%20" id="TexFormula1"

lisabon 2012 [21]

$=\dfrac{1}{x+2}\cdot\dfrac{x-5}{x-5}+\dfrac{6}{x-5}\cdot\dfrac{x+2}{x+2}\\\\=\dfrac{1(x-5)+6(x+2)}{(x+2)(x-5)}\\\\=\dfrac{7x+7}{x^{2}-3x-10}$

Usually, the "simplified" form has the polynomial multiplied out (as immediately above), rather than factored. If you want it factored, you can remove a factor of 7 from the terms of the numerator to get 7(x+1)/((x+2)(x-5)).

7 0

3 years ago