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2 years ago

Solve the simultaneous equation. 7a+3b=46 4a+5b=69

Mathematics

1 answer:

irinina [24]2 years ago

7 0

Answer:

a=1 and b=13

do you need the working out?

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The letter or number on each side indicates the side's measure and the letter or number in the interior of each angle of each tr

Travka [436]

Answer:

$\beta , \alpha , \gamma$

Step-by-step explanation:

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3 years ago

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Martin ran 6.5 miles in one hour. At that speed, how many miles will he<br> three hours?

NNADVOKAT [17]

ANSWER: Martin will run 19.5 miles in 3 hours.

STEP-BY-STEP EXPLANATION: 6.5 x 3 = 19.5

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3 years ago

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HELP ME PLEASE! Vector's map has the scaling missing.he knows from the lake to the youth center is 8 miles. on the map, they are

Naddik [55]

Hey there!

Since 2 goes into 8, you would simplify. You would get 1 inch and 4 miles. Next, set up a ratio. The ratio is 1 inch: 4 miles. So for every inch on the map, there is 4 miles.

I hope this helps!

7 0

3 years ago

Help me pls ill give extra points

olga55 [171]

Answer:

y = x + 1

Step-by-step explanation:

We are using the points (0, 1) and (6, 7). First, use these points to find the slope.

m = y₁ - y₂/x₁ - x₂

⇒ m = 7 - 1/6 - 0

⇒ m = 6/6

⇒ m = 1

Now that we know the slope is one, enter the values into the point-slope form equation. I'll use (0, 1) in this instance.

y - y₁ = m(x - x₁)

⇒ y - 1 = 1(x - 0)

⇒ y - 1 = x

⇒ y = x + 1

Therefore, the equation is y = x + 1.

3 0

3 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

3 years ago