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Sunny_sXe [5.5K]
3 years ago
10

Anybody know how to solve ??????

Mathematics
1 answer:
lys-0071 [83]3 years ago
3 0

Have you gone over sin cos and tan

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5. Original price of concert tickets:$100 . Discount: 21%
lutik1710 [3]

Answer:

Concert ticket now costs $79.

Step-by-step explanation:

21% is being removed from the original cost, in this case, that's $21.  100-21=79. Hope this helps ;)

4 0
3 years ago
Read 2 more answers
Which of the following shows the factors of 5x2 + 17x + 6? A. (5x + 1)(x + 6) B. (5x + 2)(x + 3) C. (5x + 3)(x + 2) D. (5x + 6)(
zysi [14]

Answer:

B. (x+3)(5x+2)

Step-by-step explanation:

The given quadratic trinomial is;

5x^2+17x+6

Comparing this to ax^2+bx+c, we have a=5,b=17,c=6.

ac=5\times6=30

Two factors of 30 that add up to 17 is 15  and 2.

We split the middle term to obtain;

5x^2+15x+2x+6

We factor by grouping;

5x(x+3)+2(x+3)

Factor further

(x+3)(5x+2)

The correct answer is B.

5 0
4 years ago
Read 2 more answers
Solve for x in the equation 1.23^x=10
Rzqust [24]

Answer:

x= log123/100*(10)

Step-by-step explanation:

7 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
4 years ago
Use exponents to write 81 in 3 different ways
denis23 [38]
9 to the second power 3 to the fourth power and 81 to the first power
8 0
3 years ago
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