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4 years ago

Solve the system of equations above

Mathematics

1 answer:

Law Incorporation [45]4 years ago

8 0

So, if y = 1, then all you have to do is plug in 1 for y in the first equation! :)

2x + 4y = 10

2x + 4(1) = 10

Simplify.

2x + 4 = 10

Subtract 4 from both sides.

2x = 6

Divide both sides by 2.

x = 3

So, your answer is a) x = 3, y = 1

~Hope I helped!~

You might be interested in

Two old RSM students were standing ten feet apart when they ran in opposite directions with speeds of 180 feet per minute and 16

inessss [21]

The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes

Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.

Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'

D = 180t + 10 + 160t

D = 340t + 10

<h3>Number of minutes before they are 1870 feet</h3>

Making t subject of the formula, we have

t = (D - 10)/340

Since they are 1870 feet apart after t minutes, D = 1870 feet.

t = (D - 10)/340

t = (1870 - 10)/340

t = 1860/340

t = 5.47 minutes

So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes

Learn more about minutes of distance apart here

brainly.com/question/8783264

7 0

2 years ago

6. A prism has bases that are equilateral triangles with sides lengths of 10 inches and a length of 30 inches.

eimsori [14]

Answer:

The volume of the prism is $1,299\ in^{3}$

Step-by-step explanation:

we know that

The volume of the prism is equal to

$V=BL$

where

B is the area of the triangular base

L is the length of the prism

we have

$L=30\ in$

The area of a equilateral triangle is equal to

$B=\frac{1}{2}(10)^{2} sin(60\°)$

$B=25\sqrt{3}\ in^{2}$

substitute

$V=(25\sqrt{3})(30)=1,299\ in^{3}$

8 0

3 years ago

P divided by 8????? asap nowwww

zlopas [31]

Answer:

I am not sure but if p/8=0 then it is correct

hope it helps! :D

3 0

3 years ago

Read 2 more answers

Find the value of x then classify the triangle

Sliva [168]

Answer: 88º

Step-by-step explanation:

180-47=133-45=88

4 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