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3 years ago

Solve this problem. What is x? x ÷ 7 = 6 x =

Mathematics

1 answer:

Marat540 [252]3 years ago

6 0

So another way to look at is 7x6=__ and so when you do this you get 42. Hope this is helpful.

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How many times do 14 go into 50

masha68 [24]

=3 times it goes into

4 0

2 years ago

Need help asap pls :)

pogonyaev

Answer:

See explanation below.

Step-by-step explanation:

First I'm going to find angle 2. Angle two plus 55 is equal to 115. 180-115=65. 65-55=10 Angle 2 = 10

Next, we can find angle 3. 55+10=65. 180-65=115. Angle 3 = 115

Angle 2 is equal to angle 5, angle 3 is equal to angle 6, and angle 4 is equal to 55.

Angle 5 = 10

Angle 4 = 55

Angle 6 = 115

Now we can find angle 8. 180-115=65. Angle 8 = 65

Angle 11 = 65

Angle 12 = 115

10+115=125 Angle 10 = 125

180-125 = 55 Angle 9 = 55

Angle 14 = 55

Angle 13 = 125

7 0

2 years ago

Read 2 more answers

What square root is closest to 5

nikklg [1K]

Answer: 2.236

Step-by-step explanation: The square root of 5 is expressed as √5 in the radical form and as (5)½ or (5)0.5 in the exponent form. The square root of 5 rounded up to 5 decimal places is 2.23607. It is the positive solution of the equation x2 = 5.

7 0

2 years ago

If f(x)= 3x + 4, then what is f(-2)

iren [92.7K]

Answer:

f ( - 2 ) = - 2

Step-by-step explanation:

Step 1:

f ( x ) = 3x + 4 Equation

Step 2:

f ( - 2 ) = 3 ( - 2 ) + 4 Input x value

Step 3:

f ( - 2 ) = - 6 + 4 Combine Like Terms

Answer:

f ( - 2 ) = - 2 Combine Like Terms

Hope This Helps :)

6 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