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gayaneshka [121]
3 years ago
6

Properties of multiplication 50b=10what is the answer?

Mathematics
1 answer:
poizon [28]3 years ago
8 0
B is equal to 1/5. Hope I was able to help

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Vilka [71]
I wish I had an answer for you! Unfortunately, this one has stumped even professional mathematicians for over a century. It was first presented by mathematician Edmund Landau in 1912, and it’s gone unproven ever since. I can’t give you a solution, but I can definitely say you should submit your findings to a mathematical journal if you ever find one!
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!! WILL MARK BRAINLIEST!
Lana71 [14]

Answer:

the debt ratio remains the same

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2 years ago
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
3 years ago
Use the formula S equals 2 left-parenthesis l w right-parenthesis plus 2 left-parenthesis l h right-parenthesis plus 2 left-pare
frosja888 [35]

Answer: 52 square cm

Step-by-step explanation:

The given surface area formula of a rectangular prism,

S = 2 wl + 2 lh + 2 hw

Where  is the length, w is the width and h is the height of the prism,

Here, length, l = 2 cm,

Width, w = 3 cm

And, Height, h = 4 cm

Thus,The required surface area of the prism,

S = 2 × 3 × 2 + 2 × 2 × 4 + 2 × 4 × 3

S = 12 + 16 + 24 = 52 square cm.

6 0
3 years ago
Vicente has a prism like water tank whose base area is 1.2 square meters. He bought 6 goldfish at the store , and the store owne
Reika [66]

Answer:

1.25 meters

Step-by-step explanation:

4 0
3 years ago
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