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3 years ago

What percent of 2/7 is 1/7

Mathematics

1 answer:

svlad2 [7]3 years ago

3 0

Answer:

50%.

Step-by-step explanation:

1/7 / 2/7

= 1/7 * 7/2

= 7/14

= 1/2.

= 50%.

You might be interested in

(-11+ √-4) + (4+ √-36)

mash [69]

Answer:

$-7 + 8i$

Step-by-step explanation:

$(-11 + \sqrt{-4}) + (4 + \sqrt{-36}) = (-11 + 2\sqrt{-1}) + (4 + 6\sqrt{-1})$

We know $\sqrt{-1}$ is a complex number and is defined by $i$

∴ $(-11 + \sqrt{-4}) + (4 + \sqrt{-36}) = (-11 + 2i) + (4 + 6i) = -7 + 8i$

4 0

2 years ago

Find out how much red paint in liters did they use

Lina20 [59]

Answer:

hmmmm.......

Step-by-step explanation:

love from deadpool

7 0

2 years ago

Write an explicit formula for the An, and the n^th term of the sequence 24,-12,6,....

Annette [7]

Answer:

$T_n = -48* (-\frac{1}{2})^{n}$

Step-by-step explanation:

Given

$24, -12, 6,...$

Required

Write a formula

The above sequence shows a geometric sequence because:

$r = \frac{-12}{24} = \frac{6}{-12} = -\frac{1}{2}$ -- common ratio

The equation is determined using:

$T_n = ar^{n-1}$

Where

$a = 24$

Substitute values for a and r

$T_n = 24* (-\frac{1}{2})^{n-1}$

Apply law of indices:

$T_n = 24* (-\frac{1}{2})^{n} * (-\frac{1}{2})^{-1}$

$T_n = 24* (-\frac{1}{2})^{n} * 1/(-\frac{1}{2})$

$T_n = 24* (-\frac{1}{2})^{n} * 1*-2$

$T_n = 1*-2*24* (-\frac{1}{2})^{n}$

$T_n = -48* (-\frac{1}{2})^{n}$

<em>The above represents the explicit formula</em>

7 0

2 years ago

1+4=5 2+5=12 3+6=21 puzzle

wel

Answer:

The puzzle is first add the current two numbers and then add their sum to result obtained in previous step.

Step-by-step explanation:

1 + 4 = 5

2 + 5 = 12

12 is obtained by adding 2 + 5 = 7 and adding the result obtained in previous result which is 5 to 7

⇒ 7 + 5 = 12

Now, 3 + 6 = 21

21 is obtained by first adding 3 + 6 = 9 and the previous result obtained is 12

So adding 12 to 9 ⇒ 12 + 9 = 21

Hence, The puzzle is first add the current two numbers and then add their sum to result obtained in previous step.

5 0

2 years ago

Read 2 more answers

Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,

Anton [14]

Answer:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

Step-by-step explanation:

So we have the equation:

$\tan(x-y)=\frac{y}{8+x^2}$

And we want to find dy/dx.

So, let's take the derivative of both sides:

$\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[\tan(x-y)]$

We can use the chain rule, where:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

$=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])$

Differentiate x like normally. Implicitly differentiate for y. This yields:

$=\sec^2(x-y)(1-y')$

Distribute:

$=\sec^2(x-y)-y'\sec^2(x-y)$

And that is our left side.

Right Side:

We have:

$\frac{d}{dx}[\frac{y}{8+x^2}]$

We can use the quotient rule, where:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

f is y. g is (8+x²). So:

$=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}$

Differentiate:

$=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

And that is our right side.

So, our entire equation is:

$\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

$((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)$

The right side cancels. Let's distribute the left:

$\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy$

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

$\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2$

Move -2xy to the left. So:

$\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2$

Factor out a y' from the right:

$\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)$

Divide. Therefore, dy/dx is:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}$

We can factor out a (8+x²) from the denominator. So:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

And we're done!

8 0

2 years ago