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lesya [120]
3 years ago
14

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
You might be interested in
What is (-24)-(-65)=
erica [24]

Answer:

41

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

So

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

so

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
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