1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Eva8 [605]
3 years ago
6

120% of 118 is what? percent word problems

Mathematics
2 answers:
Studentka2010 [4]3 years ago
7 0
Lets solve the problem with a rule of three, knowing that 118 is the 100%:
100% -------> 118
120% -------> x
x = (120)(118)/100
x = 141.6
therefore the 120% of 118 is 141.6
inysia [295]3 years ago
6 0
<span>118 X 120/100 =141.6 </span>
You might be interested in
How do you do this?
evablogger [386]

Answer:

7

Step-by-step explanation:

because y is square by green and 7 is also green I guess

8 0
3 years ago
Prove that if n is a perfect square then n + 2 is not a perfect square
notka56 [123]

Answer:

This statement can be proven by contradiction for n \in \mathbb{N} (including the case where n = 0.)

\text{Let $n \in \mathbb{N}$ be a perfect square}.

\textbf{Case 1.} ~ \text{n = 0}:

\text{$n + 2 = 2$, which isn't a perfect square}.

\text{Claim verified for $n = 0$}.

\textbf{Case 2.} ~ \text{$n \in \mathbb{N}$ and $n \ne 0$. Hence $n \ge 1$}.

\text{Assume that $n$ is a perfect square}.

\text{$\iff$ $\exists$ $a \in \mathbb{N}$ s.t. $a^2 = n$}.

\text{Assume $\textit{by contradiction}$ that $(n + 2)$ is a perfect square}.

\text{$\iff$ $\exists$ $b \in \mathbb{N}$ s.t. $b^2 = n + 2$}.

\text{$n + 2 > n > 0$ $\implies$ $b = \sqrt{n + 2} > \sqrt{n} = a$}.

\text{$a,\, b \in \mathbb{N} \subset \mathbb{Z}$ $\implies b - a = b + (- a) \in \mathbb{Z}$}.

\text{$b > a \implies b - a > 0$. Therefore, $b - a \ge 1$}.

\text{$\implies b \ge a + 1$}.

\text{$\implies n+ 2 = b^2 \ge (a + 1)^2= a^2 + 2\, a + 1 = n + 2\, a + 1$}.

\text{$\iff 1 \ge 2\,a $}.

\text{$\displaystyle \iff a \le \frac{1}{2}$}.

\text{Contradiction (with the assumption that $a \ge 1$)}.

\text{Hence the original claim is verified for $n \in \mathbb{N}\backslash\{0\}$}.

\text{Hence the claim is true for all $n \in \mathbb{N}$}.

Step-by-step explanation:

Assume that the natural number n \in \mathbb{N} is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number a (a \in \mathbb{N}) such that a^2 = n.

Assume by contradiction that n + 2 is indeed a perfect square. Then there should exist another natural number b \in \mathbb{N} such that b^2 = (n + 2).

Note, that since (n + 2) > n \ge 0, \sqrt{n + 2} > \sqrt{n}. Since b = \sqrt{n + 2} while a = \sqrt{n}, one can conclude that b > a.

Keep in mind that both a and b are natural numbers. The minimum separation between two natural numbers is 1. In other words, if b > a, then it must be true that b \ge a + 1.

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by (a + 1) and use the fact that b \ge a + 1 to make the left-hand side b^2.)

b^2 \ge (a + 1)^2.

Expand the right-hand side using the binomial theorem:

(a + 1)^2 = a^2 + 2\,a + 1.

b^2 \ge a^2 + 2\,a + 1.

However, recall that it was assumed that a^2 = n and b^2 = n + 2. Therefore,

\underbrace{b^2}_{=n + 2)} \ge \underbrace{a^2}_{=n} + 2\,a + 1.

n + 2 \ge n + 2\, a + 1.

Subtract n + 1 from both sides of the inequality:

1 \ge 2\, a.

\displaystyle a \le \frac{1}{2} = 0.5.

Recall that a was assumed to be a natural number. In other words, a \ge 0 and a must be an integer. Hence, the only possible value of a would be 0.

Since a could be equal 0, there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that a = 0 just won't work as in the assumption.

If indeed a = 0, then n = a^2 = 0. n + 2 = 2, which isn't a perfect square. That contradicts the assumption that if n = 0 is a perfect square, n + 2 = 2 would be a perfect square. Hence, by contradiction, one can conclude that

\text{if $n$ is a perfect square, then $n + 2$ is not a perfect square.}.

Note that to produce a more well-rounded proof, it would likely be helpful to go back to the beginning of the proof, and show that n \ne 0. Then one can assume without loss of generality that n \ne 0. In that case, the fact that \displaystyle a \le \frac{1}{2} is good enough to count as a contradiction.

7 0
3 years ago
Stephanie can make 12 baskets in 25 seconds. At this rate, how many baskets could she make in 150 seconds?
Vedmedyk [2.9K]

Answer:

72 in 150 seconds.

12/25=0.48

0.48 per second (not full baskets note)

.48x150= 72 baskets

5 0
2 years ago
What is the value of x in the isosceles trapezoid below?
Firdavs [7]
2x + (10x+24) = 180
12x+24 = 180
12x = 156
x = 13

answer x is 13
5 0
3 years ago
Read 2 more answers
The quilt is 2.44 m long and 1.83 m wide what is the area of the quilt in square feet
Alenkasestr [34]
Area= L*W

So 2.44 *1.83=4.4652

Then convert: 48.0630128
8 0
3 years ago
Other questions:
  • Find Y pls<br> Thank you
    14·2 answers
  • Can you solve for ×/4+7=5
    8·1 answer
  • Given that (2,4) is on the graph of f(x), find the corresponding point for the function f(x)-5
    14·1 answer
  • Can someone help me with these two math problems. ( Will Mark Brainliest). ​
    13·1 answer
  • Solve for x.
    7·1 answer
  • The area of a circle is 77.28m2.<br> Find the length of the diameter rounded to 1 DP.
    15·1 answer
  • Helpppppppppppppppppppppp!?!??!?!??!?!?!? Meh i need help bad it all below and i wasn't stu.pid this time a put it below like i
    10·1 answer
  • 15=-3(2c+d) solve for c
    5·1 answer
  • Find the total surface area of the prism pictured.
    14·1 answer
  • Jack Frost was decorating the car windows in the teacher’s parking lot. He puts 2 snowflakes on each of the four windows of the
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!