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3 years ago

PQRS is a parallelogram. Find the m

Mathematics

2 answers:

timurjin [86]3 years ago

8 0

Answer:

R = 70

Q = 110

Step-by-step explanation:

In a parallelogram

Opposite angles are equal

P = R = 70

Same side angles are supplementary ,which means they add to 180

P+Q =180

70+Q =180

Q = 180-70

Q = 110

viktelen [127]3 years ago

7 0

Answer:

Step-by-step explanation:

<R=<P=70degrees

<Q=180-70=110 degrees

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Is x^5 - 5x^4 + 4x^3 - 3x^2 + 2x + 1 a polynomial

vfiekz [6]

Answer:

yes

Step-by-step explanation:

Polynomial Long Division :

4.3 Polynomial Long Division

Dividing : x5-5x4+10x3-10x2+5x-1

("Dividend")

By : x-1 ("Divisor")

dividend x5 - 5x4 + 10x3 - 10x2 + 5x - 1

- divisor * x4 x5 - x4

remainder - 4x4 + 10x3 - 10x2 + 5x - 1

- divisor * -4x3 - 4x4 + 4x3

remainder 6x3 - 10x2 + 5x - 1

- divisor * 6x2 6x3 - 6x2

remainder - 4x2 + 5x - 1

- divisor * -4x1 - 4x2 + 4x

remainder x - 1

- divisor * x0 x - 1

remainder 0

Quotient : x4-4x3+6x2-4x+1 Remainder: 0

4 0

3 years ago

Whoever answers CORRECTLY will get Brainliest. Do not answer if you don’t know. Don’t guess either. I want all correct answers.

nignag [31]

Answer: 18 cm^2

Step-by-step explanation:

There are four equal triangle and a rectangle.

Formula of a rectangle = b * h = 2*3 = 6

Formula of triangle = $\frac{bh}{2}$ but there four of them so we times them by 4, this leaves us with 2 (b * h) = 2(2*3) = 12

Area of pentagon = area of all triangles + rectangle

Area of pentagon = 12 + 6 = 18

Hope this helps.

4 0

3 years ago

Read 2 more answers

Why do banks charge fees?

Ivahew [28]

They charge fees so they can cover there costs and return value to shareholders

3 0

3 years ago

What are the roots of the polynomial equation?

Andreas93 [3]

Answer:

The roots of the polynomial equation in this case would be the intersection of the 2 polynomial functions. which are at x = 4 and x = -3

Step-by-step explanation:

The roots are found by finding the x-values of the intersections of these two cubic polynomial functions.

We could try solving algebraically, but you have the graph.

6 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 <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