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Bingel [31]
3 years ago
12

Need some help please :)

Mathematics
1 answer:
aivan3 [116]3 years ago
5 0

Answer:

B) No

Step-by-step explanation:

10²+15²is not 20²

so its B

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If t1 = 4, s1 = 5, and s2 = 2, determine the value of t2
makkiz [27]
The domain is the set { -2, 0, 2, 3}
discussion:
the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives -2, 0, 2, 3.
6 0
2 years ago
10<br> 10<br> O A. (3,8)<br> B. (6,5)<br> O c. (5,6)<br> O D. (8,3)
Svetradugi [14.3K]

Answer:

B.(6,5)

Step-by-step explanation:

because (6,5) ang sagot

8 0
3 years ago
1. The sum of a two-digits number is 13. The tens digit is 8 less than twice the units digit. What is the number?
alexgriva [62]

Answer:

<u>The number is 67</u>

Step-by-step explanation:

<u>Equations</u>

Let's consider the number 83. The tens digit is 8 and the unit digit is 3. Note the tens digit's addition to the number is 80, and the unit's addition is 3. This means the tens digit adds 10 times its value, that is, 83 = 8*10 + 3.

Now, let's consider the number ab, where a is the tens digit, and b is the unit digit. It follows that

Number=10*a+b

The question gives us two conditions:

1) The sum of a two-digits number is 13.

2) The tens digit is 8 less than twice the units digit.

The first condition can be expressed as:

a + b = 13                     [1]

And the second condition can be written as:

a = 2b-8                      [2]

Replacing [2] into [1], we have:

2b-8 + b = 13

Operating:

3b = 13 + 8

3b = 21

Solving for b:

b = 21 / 3 = 7

Substituting into [2]:

a = 2*(7) - 8 = 6

Thus, the number is 67

3 0
3 years ago
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
3. Michel invests $850 at 7%/a simple interest. How long will he
Kipish [7]

~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill&\$200\\ P=\textit{original amount deposited}\dotfill & \$850\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ t=years \end{cases} \\\\\\ 200=850(0.07)t\implies \cfrac{200}{850(0.07)}=t\implies \stackrel{\textit{about 3 years, 4 months and 10 days}}{3.36\approx t}

5 0
2 years ago
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