All categories

3 years ago

Need help asap plzzz

Mathematics

2 answers:

Anna11 [10]3 years ago

7 0

Irrational. The answer is “irrational”.

Scilla [17]3 years ago

4 0

$\sqrt{12}$
enters the irrational number category.

Integers numbers are the numbers without decimals between -infinity and infinity.

Natural numbers are the whole numbers from 1 to infinity.

Whole numbers are natural numbers + 0.

You might be interested in

What is the area of this figure?

irakobra [83]

40+25+25+70=160 Hope this helps! :)

6 0

3 years ago

Read 2 more answers

This equation shows how the revenue at Jill's Auto Wash depends on the number of

Romashka [77]

Answer:y = 9x

Step-by-step explanation:

6 0

2 years ago

Cedric completed the square for the quadratic expression 5x^2−40x+15 in order to determine the minimum value of the expression,

Olegator [25]

Answer:

Step 3 is not correct (should be $(x-4)^2$ , not $(x+4)^2$

Step-by-step explanation:

The expression that we have to simplify is:

$5x^2-40x+15$

We proceed step-by-step, in order to find the mistake did by Cedric:

1) First, we factorize the 5 outside:

$5(\frac{5x^2}{5}-\frac{40x}{5}+\frac{15}{5})=5(x^2-40x+3)$ --> this step is correct

2) We add +16 and -16 inside the brackets:

$5(x^2-8x+16-16+3)$ --> this step is correct

3) We rewrite the term $(x^2-8x+16)$ as the square of a bynomial, which is $(x-4)^2$ , so the expression becomes

$5((x-4)^2-16+3)$ --> we notice that this step is wrong: in fact, Cedric wrote $(x+4)^2$ , which is not correct.

4) Now we continue: we rewrite -16+3 as -13,

$5((x-4)^2-13)$

5) Finally, we multiply the 5 by the terms in the bracket:

$=5(x-4)^2-65$

5 0

3 years ago

What basic trigonometric identity would you use to verify that

Dafna1 [17]

Given the equation:

$\frac{\sin^2x+\text{cos}^2x}{\cos x}=\sec x$

Let's determine the trigonometric identity that you could be used to verify the exquation.

Let's determine the identity:

Apply the trigonometric identity:

$\sin ^2x+\cos ^2x=1$

$\cos x=\frac{1}{\sec x}$

Replace cosx for 1/secx

Thus, we have:

$\begin{gathered} \frac{\sin^2x+\cos^2x}{\frac{1}{\sec x}} \\ \\ =(\sin ^2x+\cos ^2x)(\sec x) \\ \text{Where:} \\ (\sin ^2x+\cos ^2x)=1 \\ \\ We\text{ have:} \\ (\sin ^2x+\cos ^2x)(\sec x)=1\sec x=\sec x \end{gathered}$

The equation is an identity.

Therefore, the trignonometric identity you would use to verify the equation is:

$\cos ^2x+\sin ^2x=1$

ANSWER:

$\cos ^2x+\sin ^2x=1$

6 0

1 year ago

Please help me solve question 5.) with the letters a) & b)

tatyana61 [14]

Answer:

Step-by-step explanation:

5.(a)

x²+6x-15=0

$x=\frac{-6\pm\sqrt{36+4*15}}{2*1}\\x=\frac{-6\pm\sqrt{36+60} }{2}\\x=\frac{-6\pm\sqrt{96} }{2}\\x=\frac{-6\pm4\sqrt{6} }{2}\\x=-3\pm2\sqrt{6}$

(b)

4x²-8x-1=0

$x=\frac{8\pm\sqrt{64+4*4}}{2*4}\\x=\frac{8\pm\sqrt{80}}{8}\\x=\frac{8\pm 4\sqrt{5} }{8}\\x=\frac{2\pm\sqrt{5}}{2}$

7 0

3 years ago