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kirill115 [55]
3 years ago
16

The function f(x) = (10)–x is reflected across the x-axis to create the function g(x). Which ordered pair is on g(x)? (–2, 75) (

3, –750)
Mathematics
2 answers:
enot [183]3 years ago
7 0

Answer:

(2, -3/400)

Step-by-step explanation:

Leto [7]3 years ago
4 0
The correct answer is C) <span>(2, -3/400) Hope this helps!!!!</span>
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I still haven't finished this and it's due today. i really would appreciate some help .
Svetlanka [38]
A
x X x
= x^2

B
20 x 20 = 40

C
100 / 40 = 2.5 per square inch
4 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
a guy wire is attached to a tree 3.5ft above the ground to stabilize it. if the guy wire forms an angle with the tree of 50°, wh
balandron [24]
The plane figure formed by the ground, the guy wire, and the tree is a right triangle with the hypotenuse equal to length of the guy wire. The angle given is an angle adjacent to 3.5 ft. Therefore, the most suitable trigonometric function for this is,
                                  cos (50°) = adjacent / hypotenuse
                                   cos 50° = 3.5 ft / hypotenuse
The value of the hypotenuse is 5.445 ft. 
Hence, the length of the guy wire is approximately 5.445 ft. 
4 0
3 years ago
What equation is (x+4)^2 equal to
elena55 [62]

Answer: x^2+8x+16

Step-by-step explanation:

Following (a+b)^2=a^2+2ab+b^2,

(x+4)^2=x^2+2(4x)+(4)^2

x^2+8x+16

3 0
3 years ago
Which graph correctly represents 1/3y-1/2x&gt;2 ?
Tanzania [10]
<span>Put it in the form of y =mx +b, or in this instance, y> mx +b 
move the (1/2) x to the right by adding it to both sides of the inequality 
(1/3)y>(1/2)x +2 
Multiply by 3 on both sides to get y by itself. 
y>(3/2) x +6 
This is a graph with y intercept of (0,6) and a moderate upward and to the right slope. Because it is > , the line on the graph will NOT be part of the solution. 

The easiest way to find the side of the graph that the inequality satisfies is to use (0,0) and see if it works or doesn't work. In the original equation, 0-0>2 does NOT work, so the area where the inequality works is to the up and left of the graph, which should be a dotted line to show that the inequality is greater than only. 
The point (6,-2) should work.
Test it. 6*(1/3)-(-2)*(1/2)>0 ; 2-(-1)=3, and 3>2 It does work. 
The point (6,2) should not work 
Test it. 6 *(1/3)-2(1/2)=2-1 ; 1 is NOT >2, so it does not work. 
If the graph goes through the origin, then pick a point near the graph with a small x or y.</span>
7 0
3 years ago
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