All categories

3 years ago

Please help me as i dont quite uinderstand

Mathematics

2 answers:

amid [387]3 years ago

6 0

Answer:

sorry i don't know

Step-by-step explanation:

but i can help you with anything else

just give me a message and i will be straight with you

Rzqust [24]3 years ago

4 0

Answer:

Step-by-step explanation:

Using the rules of exponents/ radicals

$a^{-m}$ ⇔ $\frac{1}{a^{m} }$

$(^m)^{n}$ = $a^{mn}$

$a^{\frac{m}{n} }$ ⇔ $\sqrt[n]{a^{m} }$

Thus

$(\frac{1}{27}) ^{\frac{1}{3} }$ = $\frac{1}{27^{\frac{1}{3} } }$ = $\frac{1}{\sqrt[3]{27} }$ = $\frac{1}{3}$

$(\frac{1}{81}) ^{-\frac{3}{4} }$

= $\frac{1}{81^{-\frac{3}{4} } }$

= $81^{\frac{3}{4} }$

= $(\sqrt[4]{81} )^3$ = 3³ = 27

Thus

$\frac{1}{3}$ × 27 = 9

You might be interested in

Can someone help me I give you 10 points

natka813 [3]

Answer:

A:40%

B:7.8

Step-by-step explanation:

Using a proportion , you can solve for the bottom side.

13/19.5 = 5.2/x

multiply 19.5 and 5.2 and rewrite the equation

101.4=13x

x=7.8

To find the scale, divide 5.2 by 13

5.2/13=0.4

Double check by dividing 7.8 by 19.5

7.8/19.5=0.4

0.4=40%

7 0

3 years ago

.If x= 2 +√3 , find the value of x^2+1/x^2 and x^4+1/x^4

kkurt [141]

Answer:

pls may you explain I don't understand explain step by step

5 0

3 years ago

2-2x<4? Can someone please help me!

yKpoI14uk [10]

Answer:

x > -1

Step-by-step explanation:

To solve the inequality, use the same rules as solving equations.

2 - 2x < 4

-2x < 2

x > -1

Since you divide by a negative, you must switch the sign < to >.

6 0

3 years ago

2 5/12 x 2 2/3 pls helpp

MA_775_DIABLO [31]

Answer: 58/9

Step-by-step explanation:

5 0

3 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using reductio ad absurdum. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by reductio ad absurdum that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

4 years ago