All categories

2 years ago

I need help with this questionn

Mathematics

1 answer:

hichkok12 [17]2 years ago

7 0

56,,, I got that by plugging in the numbers from the right triangle and used the Pythagorean theorem.

You might be interested in

No Explanation Please Don't Answer

vazorg [7]

Answer:

A) $\frac{1}{2}(9+15)w$

Step-by-step explanation:

Area of a trapezoid is $\frac{1}{2}(base1+base2) *h$

$A=\frac{1}{2}(9+15)*w$

8 0

2 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 <em>reductio ad absurdum</em>. 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 <em>reductio ad absurdum</em> 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

3 years ago

When will the tuition reach $20,000 according to the trendline?

Maru [420]

Answer:

2050

Step-by-step explanation:

3 0

2 years ago

82. Close shave Which of two brands (X or Y) of electric

KiRa [710]

Yes make. lavas labs ova w labs. alan s slabs. sks labs. siena. alana. alana w is a whelks d fndkrn darkens kebab w eke. sm she. sm sm sm s is sks s sm s sm sm s sm. and. sm sks dinks sks sks sj she s she when w

7 0

3 years ago

The product of b and 2 is greater than or equal to 28

dimaraw [331]

Answer:

$b \geqslant 14$

Step-by-step explanation:

Product of b and 2 = 2b

According to the given description:

$2b \geqslant 28 \\ \\ b \geqslant \frac{28}{2} \\ \\ b \geqslant 14$

4 0

2 years ago