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3 years ago

15 Points!!! Will Give Brainliest!!! Help Plzz!!!

2 answers:

8 0

C is the answer because its the only one that makes sense and matches the following statement you have in your question.

Luden [163]3 years ago

4 0

Answer:

Step-by-step explanation:

Converse. Switching the hypothesis and conclusion of a conditional statement. For example, the converse of "If it is raining then the grass is wet" is "If the grass is wet then it is raining."

Note: As in the example, a proposition may be true but have a false converse.

So basically you just switch the answers around, to give you the same meaning, but it a different format.

Hope this helps, sorry if not, have a good day. c;

A lot of these make sense to me, but I think this one is the best answer.

You might be interested in

Write a system of equations in x and y describing the situation. Do not solve the system. Keiko has a total of $ 5600 comma$5600

IgorLugansk [536]

Answer:

The correct answers are x + y = 5600 and x - y = 700.

Step-by-step explanation:

Write a system of equations in x and y describing the situation. Do not solve the system.

Keiko has a total of $ 5600, which she has invested in two accounts.

Let x be the amount of money in the larger account and y be the amount of money in the smaller account.

So, Keiko has invested her $5600 in two accounts x and y.

Thus x + y = 5600.

Also, given by the problem, the larger account (x) is $ 700 greater than the smaller account (y).

Thus x - y = 700.

Thus the two system of equation in x and y describing the given situation are

x + y = 5600 and x - y = 700

8 0

3 years ago

Solve 3(4m - 6) = 12 and explain each step

Studentka2010 [4]

Distribute the 3 to 4m and 6. this makes 12m-18=12
now add 18 to both sides.
12m=30
now divide by 12
m=30/12

3 0

3 years ago

The function, f(x)=70n-400, models the profit of the instructor of a guitar class per month, where n is the number of students e

zimovet [89]

profit=f(x)

so solve for f(x)=1000

1000=70n-400

add 400 to both sides

1400=70n

divide both sides by 70

20=n

20 students need to be enrolled to make a profit of 1000.00

4 0

3 years ago

Please Help!!!<br><br> What is a polynomial function in standard form with zeroes 1, 2, -3, and -1 ?

AlekseyPX

Answer:

Numeric expressions apply operations to numbers. For example, 2(3 + 8) is a numeric expression. Algebraic expressions include at least one variable and at least one operation (addition, subtraction, multiplication, division). For example, 2(x + 8y) is an algebraic expression.

Step-by-step explanation:

7 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

4 years ago