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Dennis_Churaev [7]
3 years ago
11

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
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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
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