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3 years ago

Answer the question based on the data in the two-way table.

Mathematics

2 answers:

kirill [66]3 years ago

6 0

Answer:

...

Step-by-step explanation:

umm... honestly we will NEVER use this in the real world ‍♂️

liubo4ka [24]3 years ago

5 0

The answer is probably c

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What is (-24)-(-65)=

erica [24]

Answer:

Step-by-step explanation:

=(-24)-(-65)

= -24 + 65

= 65-24

= 41

8 0

3 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

3 years ago

Math graph equation. <br> (Picture.)

vladimir1956 [14]

Answer:

y = -2x + 3 is the required equation

Step-by-step explanation:

From the graph we will get two ordered pairs

If we see that When x=0 then y=3 and when x = 1 then y=1

Now from this we can find the slope of the Graph

So slope of graph = (y2 - y1) / (x2-x1)

=(1-3) / (1-0)

=-2

Slope = m = -2

Now the equation is

y = mx + b

We have to find the value of b

as we have two ordered pairs we can take any of it to find the value of b because line passes through those points

taking (0,3) as a point

y= mx + b

put values

3 = -2*(0) + b

b=3

Now the equation of the line in this form is

y = mx + b

which is

y = -2x + 3

5 0

3 years ago

12.917 rounded to the nearest ones

adelina 88 [10]

I am pretty sure the answer is 13

6 0

3 years ago

Read 2 more answers

What is the value of 5^3i^9

Tema [17]

Answer:

5^(3)i^(9)= five cubes times i to the power of 9=125i

Step-by-step explanation:

Raise 5 to the power of 3

rewrite i9 as (i^4)2^i

so, you get 125((i^4)^2i)

i^4=1

125(1^2i)

125i

4 0

3 years ago