All categories

2 years ago

Help pleaseeeee????????

Mathematics

2 answers:

rodikova [14]2 years ago

3 0

Answer is C sir......

Alexeev081 [22]2 years ago

3 0

The answer to this question is B.

You might be interested in

Suppose the leader of a camping trip is putting together a trail mix

wel

Answer:

8 pounds of peanuts, 6 pounds of raisins and 6 pounds of chocolate chips

Step-by-step explanation:

Let x be the number of pounds of peanuts and y be the number of pounds of raisins and chocolate chips.

Peanuts cost $2 per pound, then x pounds cost $2x.

Raisins cost $2.50 per pound, then y pounds cost $2.50y.

Chocolate chips cost $4 per pound, then y pounds cost $4y.

In total, x+y+y=20 and those 20 pounds cost

2x+2.50y+4y=20·2.75.

Solve the system of two equations:

$\left \{ {{x+2y=20} \\ \\ \\ \\ \\ \\ \atop {2x+6.5y=55}} \right.$

From the first equation:

$x=20-2y$

Substitute x into the second equation:

$2(20-2y)+6.5y=55\\ \\40-4y+6.5y=55\\ \\2.5y=15\\ \\25y=150\\ \\y=6\\ \\x=20-2\cdot 6=8$

8 0

3 years ago

Need Help with #5<br> 4 pts

olga_2 [115]

Answer = 10=21

When you’re multiplying numbers with exponents with the same base (10 is the base) all you have to do is add the exponents

18 and 3 are the exponents we’re dealing with, so we add them and get 21. The base stays the same

10^18 x 10^3 = 10^21

Answer = 10=21

8 0

3 years ago

Write in point-slope form an equation of the line that passes through the point (8, 9) with slope 7.

Lina20 [59]

Answer:

Step-by-step explanation:

$(8,9)=(x1,y1) \\ m=7 \\ y−y1=m(x−x1) \\ y−9=7(x−8)$

$y - 9 = 7x - 56 \\ y = 7x - 56 + 9 \\ y = 7x - 47$

8 0

3 years ago

A the fraction one half kilogram of flour is packed equally into 8 bags. How much flour is in each bag

marissa [1.9K]

Answer:

0.0625 kg per bag

or 1/16th of a kilogram per bag

Step-by-step explanation:

You'd need to divide the total amount the the number of sections.

In this case, it'd be 0.5 kg/8 bags, resulting in 0.0625 kg of flour per bag.

I hope this helps!

6 0

2 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

3 years ago