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Tom [10]
4 years ago
12

Consider the statement "There are no simple solutions to life’s problems." Write an informal negation for the statement, and the

n write the statement formally using quantifiers and variables.

Mathematics
2 answers:
Natasha_Volkova [10]4 years ago
4 0

Answer:

Step-by-step explanation:

To write the informal negation to the statement.

Let us first consider a statement and what the negation of a statement is defined as.

A statement is any sentence that can be determined as true or false. A statement can be written as P(x) or Q(x) .

For example, P(x) = 'a' likes rices.

Q(x) = b is a Maths student.

A negation to a statement is the opposite to that statement.

The negation of a statement P(x) is written as ~P(x).

If P(x) is "I am a boy", then ~P(x) (not P(x)) is "I am not a boy"

The negation of “There are no simple solutions to life’s problems” is “There are simple solutions to life’s problems.”

To write the statement formally using quantifiers.

MATHEMATICAL QUANTIFIERS:

1. For all, denotes as ∀

2. There exists, denoted as ∃

Let us define a Domain and a Set. A Set is simply a collection of objects.

Let X be the set of all simple solutions to all problems.

∀ x ∈ X, x is not a solution to life's problems.

Blizzard [7]4 years ago
3 0

solution is in the attachment

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Takuya's parents gave him \$100$100dollar sign, 100 as a birthday gift. Since he loves board games, he spent \$20$20dollar sign,
Delicious77 [7]

Answer:

The amount of money that remained from Takuya's present during month n is expressed as Tn = 120-20n and the sequence is an ARITHMETIC SEQUENCE.

Step-by-step explanation:

If Takuya's parents gave him $100 as a birthday gift and she spent $20 each month on board games until his money ran out, this means he keeps spending $20 every month and his money keep reducing by the same amount each month until his money ran out, the following can be inferred;

initial amount = $100

If he spends $20 each month,

balance at the end of 1st month = $100-$20 = $80

balance at the end of 2nd month = $80-$20 = $60

balance at the end of  3rd month = $60-$20 = $40 and so on

The sequence formed by his balances is $100, $80, $60, $40...

Since the amount keep reducing by the same value i.e $20, then the sequence formed is an ARITHMETIC SEQUENCE.

The nth term of an arithmetic sequence is expressed as Tn = a+(n-1)d

a is the first term of the sequence = 100

d is the common difference = 80-100 = 60-80 = 40-60 = 20

n is the number of terms

Substituting the given values in the formula we have;

Tn = 100+(n-1)*-20

Tn = 100+(-20n+20)\\Tn = 100-20n+20\\Tn = 120-20n

The amount of money that remained from Takuya's present during month n is expressed as Tn = 120-20n and the sequence formed is an ARITHMETIC SEQUENCE

6 0
4 years ago
Wo do you find the value when solving an equation like this 5!
Brilliant_brown [7]
For 5! you can just put it in a calculator of any sort but if you want to to the work you multiply every number before it like 5*4*3*2*1 and this equals 120.
8 0
4 years ago
What expression is equivalent to 5y^-3?
sineoko [7]

Answer:

\frac{5}{y^{3}}

Step-by-step explanation:

apply exponent rule:

a^{-b}=\frac{1}{a^b}

7 0
3 years ago
West of a city, a certain eastbound route is straight and makes a steep descent toward the city. The highway has a 11% grade, wh
creativ13 [48]

Answer:

The change in horizontal distance is 1.72 mi

Step-by-step explanation:

From the question, the highway has a 11% grade, which means that its slope is -11 / 100.

The negative sign indicates that the road is descending.

Hence, slope (m) = 11/100

The slope m, is given by

Slope = \frac{Rise}{Run} = \frac{\Delta y}{\Delta x}

Where Δy is the change in vertical distance and

Δx is the change in horizontal distance

Now, from the question, you have descended a distance of 1000 ft, that is

Δy = 1000 ft

Then, to determine the change in the horizontal distance Δx,

From

Slope = \frac{\Delta y}{\Delta x}, then

\frac{11}{100} = \frac{1000ft}{\Delta x}

11\Delta x = 1000ft \times 100

\Delta x = \frac{100000ft}{11}

∴ \Delta x = 9090.91 ft

Δx = 9090.91 ft

This is the change in horizontal distance in ft, Now to convert it to miles

1 mile = 5280 ft

Hence, 1 ft = 1/5280 miles

If 1 ft = 1/5280 miles

Then, 9090.91 ft = x miles

x = 9090.91 × 1/5280

x = 1.72 miles

Hence, the change in horizontal distance in miles is 1.72 mi

4 0
4 years ago
How would you also do this??
Vesna [10]

4(5x + 7) = 128

First, I'd do distributive property and get:

20x + 28 = 128

Then, subtract 28 from each side.

20x = 100

Next, I'd divide 100 by 20 = 5

x = 5

4 *(5(5) + 7) = 128

5 * 5 = 25

4 * (25 + 7)

4 * 32 = 128

Therefore, the perimeter = 2l + 2w

l = 4

w = 5x + 7

BUT we know that x = 5

w = 5(5) + 7 = 32

2(4) + 2(32)

8 + 64 = 72

Therefore, the perimeter would be 72 inches.

5 0
4 years ago
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