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3 years ago

What this question?????????

Mathematics

2 answers:

NeTakaya3 years ago

4 0

Answer:

The Answer is A

Step-by-step explanation:

Tanzania [10]3 years ago

3 0

Answer:

A. $\frac{3}{4} = \frac{21}{28}$

Step-by-step explanation:

$\frac{21}{28}$ ÷ $\frac{7}{7} = \frac{3}{4}$

21 ÷ 7 = 3

28 ÷ 7 = 4

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7. If F(x) = f(g(x)), where f(-2) = 8, f'(-2) = 4, f'(5) = 3, g(5)=-2, gʻ(5) = 6.. Find F'(5)

serious [3.7K]

Answer:

I'm going to lay this out in a chart so it's a little easier to see:

F(x) = f(g(x))

x | f (x) | f ' (x) | g (x) | g ' (x)

--------------------------------------

-2 | 8 | 4 |

5 | | 3 | -2 | 6

Remember the chain rule, which says

(f (g (x))) ' = g ' (x) f ' (g (x))

When they ask for F ' (5), they are asking for (f (g (x))) ' when x = 5.

Using the chain rule, that's

F ' (5) = g ' (5) f ' (g (5))

We can simplify using the numbers provided.

F ' (5) = (6) f ' (-2)

F ' (5) = (6) (4)

F ' (5) = 24

I hope that helps!

by jannat <33

7 0

2 years ago

Read 2 more answers

AB=2y+1, BC=y+1,CD=7x-3,DA=3x what is x and y

sattari [20]

I attached the picture associated with this question.

Answer:
x = 2
y = 5

Explanation:
ABCD is a parallelogram. This means that each two opposite sides are equal.
This means that:
1- AB = CD
2y + 1 = 7x - 3 ...........> equation I
2- AD = BC
3x = y + 1
This can be rewritten as:
y = 3x - 1............> equation II

Substitute with equation II in equation I and solve for x as follows:
2y + 1 = 7x - 3 ...........> equation I
2(3x - 1) + 1 = 7x - 3
6x - 2 + 1 = 7x - 3
6x - 1 = 7x - 3
7x - 6x = -1 + 3
x = 2

Substitute with x in equation I to get y as follows:
y = 3x - 1
y = 3(2) - 1
y = 6 - 1
y = 5

Hope this helps :)

3 0

2 years ago

A regression analysis between weight (y in pounds) and height (x in inches) resulted in the following least squares line: ŷ = 13

Lunna [17]

Answer:

$y=6x +135$

And for this case the interpretation for the slope would be that for every unit that the height in inches increase then the weight in pounds increase 6 units.

For the intercept of 135 represent the amount initial amount of weight for the scale.

Step-by-step explanation:

We assume that they use least squares in order to create the regression equation

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

And the slope would be:

$m=6$

The means for x and y are given by:

$\bar x= \frac{\sum x_i}{n}$

$\bar y= \frac{\sum y_i}{n}$

And we can find the intercept using this:

$b=\bar y -m \bar x$

For this case we know that the line adjusted is:

$y=6x +135$

And for this case the interpretation for the slope would be that for every unit that the height in inches increase then the weight in pounds increase 6 units.

For the intercept of 135 represent the amount initial amount of weight for the scale.

3 0

3 years ago

Identify the correct measure of the Thumbtack to the nearest 1/8 inches

bulgar [2K]

Answer:

3/4 or 6/8

Step-by-step explanation:

the thumbtack goes to 12/16 on the ruler which can be simplified to 3/4 or 6/8.

6 0

3 years ago

Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.

Sedaia [141]

$\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ 
\mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ 
\mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation 
becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} 
\mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} 
\end{array}$

$\large\begin{array}{l} \textsf{Using 
the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ 
\mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ 
\mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ 
\mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ 
\mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot
 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}$

$\large\begin{array}{l}
 \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ 
\mathsf{\Delta=(4.8)^2}\\\\\\ 
\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ 
\mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ 
\mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! 
2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} 
\end{array}$

$\large\begin{array}{l} \begin{array}{rcl} 
\mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ 
\mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} 
\end{array}$

$\large\begin{array}{l} \textsf{Both 
are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ 
\begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or 
}~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse 
tangent function:}\\\\ \begin{array}{rcl} 
\mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ 
\mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ 
\textsf{where k in an integer.} \end{array}$

__________

$\large\begin{array}{l}
 \textsf{Now, restrict x values to the interval 
}\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ 
\begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} 
\end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{
 is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 
4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} 
\end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} 
\end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}$

$\large\begin{array}{l}
 \textsf{Solution set:}\\\\ 
\mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}}
 \end{array}$

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$\large\textsf{I hope it helps.}$

Tags: trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function

6 0

3 years ago