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Kisachek [45]
3 years ago
10

Which statements are true about a perfect cube monomial and its roots? Check all that apply.

Mathematics
2 answers:
GuDViN [60]3 years ago
7 0

Answer:

A E and F

Step-by-step explanation:

tangare [24]3 years ago
3 0

Answer:

A, D, F

Step-by-step explanation:

Just took the test

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What is the answer to this math equation? <br><br> -3y &lt; -14
Illusion [34]

Answer:

y = 5

Step-by-step explanation:

-3 * 5 = -15

-15 < -14

5 0
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
. Given ????(5, −4) and T(−8,12):
damaskus [11]

Answer:

a)y=\dfrac{13x}{16}-\dfrac{129}{16}

b)y = \dfrac{13x}{16}+ \dfrac{37}{2}

Step-by-step explanation:

Given two points: S(5,-4) and T(-8,12)

Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!

Perpendicular to ST:

the equation of any line is given by: y = mx + c where, m is the slope(also known as gradient), and c is the y-intercept.

to find the perpendicular of ST <u>we first need to find the gradient of ST, using the gradient formula.</u>

m = \dfrac{y_2 - y_1}{x_2 - x_1}

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)

m = \dfrac{12 - (-4)}{(-8) - 5}

m = \dfrac{-16}{13}

to find the perpendicular of this gradient: we'll use:

m_1m_2=-1

both m_1and m_2 denote slopes that are perpendicular to each other. So if m_1 = \dfrac{12 - (-4)}{(-8) - 5}, then we can solve for m_2 for the slop of ther perpendicular!

\left(\dfrac{-16}{13}\right)m_2=-1

m_2=\dfrac{13}{16}:: this is the slope of the perpendicular

a) Line through S and Perpendicular to ST

to find any equation of the line all we need is the slope m and the points (x,y). And plug into the equation: (y - y_1) = m(x-x_1)

side note: you can also use the y = mx + c to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

(y - y_1) = m(x-x_1)

we have the slope of the perpendicular to ST i.e m=\dfrac{13}{16}

and the line should pass throught S as well, i.e (5,-4). Plugging all these values in the equation we'll get.

(y - (-4)) = \dfrac{13}{16}(x-5)

y +4 = \dfrac{13x}{16}-\dfrac{65}{16}

y = \dfrac{13x}{16}-\dfrac{65}{16}-4

y=\dfrac{13x}{16}-\dfrac{129}{16}

this is the equation of the line that is perpendicular to ST and passes through S

a) Line through T and Perpendicular to ST

we'll do the same thing for T(-8,12)

(y - y_1) = m(x-x_1)

(y -12) = \dfrac{13}{16}(x+8)

y = \dfrac{13x}{16}+ \dfrac{104}{16}+12

y = \dfrac{13x}{16}+ \dfrac{37}{2}

this is the equation of the line that is perpendicular to ST and passes through T

7 0
3 years ago
PLEASE HURRY!! The graph shows the distance traveled by two cars over several minutes.
tamaranim1 [39]
Im late but i had this test its car A < rate of car B hope it helps
6 0
2 years ago
Read 2 more answers
Reflect the point (2, -4) over the y-axis. Question 1 options: (2, 4) (-2, -4) (-4, 2) (-2, 4) Question 2 will mark the brainlie
o-na [289]

Answer:

The coordinates of the image point will be (-2,-4).

Step-by-step explanation:

The reflection of the point (2,-4) over the y-axis is to be determined.

Now, as the reflecting mirror is the y-axis, then the y-coordinate of the reflecting point will not change.

So, the image point will have coordinates (h,-4).

Now, after the reflection over the y-axis, the x-coordinate of the image point will change the sign only compared to the original point.

Therefore, the coordinates of the image point will be (-2,-4). (Answer)

3 0
2 years ago
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