4<br> Solve the system of equations using the substitution method.<br> y = 2<br> 2x+y=8

Simplify the expression. 8w + 5r – t + 9r

Wittaler [7]

Answer:

14r−t+8w

Step-by-step explanation:

Let's simplify step-by-step.

8w+5r−t+9r

=8w+5r+−t+9r

Combine Like Terms:

=8w+5r+−t+9r

=(5r+9r)+(−t)+(8w)

=14r+−t+8w

Answer:

=14r−t+8w

3 0

2 years ago

. You want to buy a book worth Php 220.50. You can buy it only by making a sacrifice of not buying a soft drink every day. If so

kherson [118]

Answer:

B-14

Step-by-step explanation:

Divide 220.50 by 15.75 and you will get 14

Hope it can help you lovelots

4 0

3 years ago

Read 2 more answers

I need the answers for both a and b so thats why 61 points are up for grab

kodGreya [7K]

Answer:

Step-by-step explanation:

From the picture attached,

a). Triangle in the figure is ΔBCF

b). Since, $L_1$ and $L_2$ are the parallel lines and m is a transversal line,

m∠FBC = m∠BFG [Alternate interior angles]

Since, $L_1$ and $L_2$ are the parallel lines and n is a transversal line,

m∠BCF = m∠CFE [Alternate interior angles]

By triangle sum theorem in ΔBCF

m∠FBC + m∠BCF + m∠BFC = 180°

From the properties given above,

m∠BFG + m∠CFE + m∠BFC = 180°

m∠GFE = 180°

Therefore, angle GFE is the straight angle that will be useful in proving that the sum of the measures of the interior angles of the triangle is 180°.

5 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

Using words, how do you solve for x? <br> -20 = x + 16

nadya68 [22]

-20 = x + 16.....subtract 16 from both sides
-20 - 16 = x....combine like terms
-36 = x

7 0

3 years ago

Read 2 more answers

4 Solve the system of equations using the substitution method. y = 2 2x+y=8