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3 years ago

Easy question it’s due in 5 minutes

Mathematics

2 answers:

alexdok [17]3 years ago

7 0

Answer:

your answer is 11 so B.

zysi [14]3 years ago

4 0

I believe the Answer is B, 11

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Find the area of this shape

Korolek [52]

Answer:

1496π square meters

Step-by-step explanation:

A=2πrh+2πr²

=2π(17)(27)+2π(17)²

= 918π + 578π

=1496π

6 0

3 years ago

"Based on the following construction, which of the following are true? choose all that apply."

umka21 [38]

From the diagram,

Angle

Bisecting an angle means dividing the angle into two equal part.

1/2 m

$\begin{gathered} \text{ } \\ \\ \end{gathered}$

4 0

1 year ago

Which statement best describes the area of triangle abc shown below?

Archy [21]

Answer:

I think the fourth one and what is going on with it being upside down lol??????????????

Step-by-step explanation:

lmk if I helped but if I did not I will give ur points back:).

3 0

3 years ago

Use graphing technology to find the range of the function f(x)=-x^2+4.f(x)=−x <br> 2<br> +4.

Andre45 [30]

Answer:

The set of Real numbers

Step-by-step explanation:

x² + 4 is a parabola with the lowest point on the y-axis is 4 and continues up

-x² + 4 is a parabola with the highest point on the y-axis is 4 and continues down.

The range is the distance traveled on the y- axis.

All real numbers.

5 0

2 years ago

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the resulting values to the corresponding l

belka [17]

The correct solution to the limits of x in the tiles can be seen below.

$\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }$ $\mathbf{ = -\dfrac{1}{52} }$

$\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }$ $\mathbf{=\dfrac{1}{79} }$

$\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }$ = -31

$\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }$ $\mathbf{ =\dfrac{1}{98}}$

<h3>What are the corresponding limits of x?</h3>

The limits of x approaching a given number of a quadratic equation can be determined by knowing the value of x at that given number and substituting the value of x into the quadratic equation.

From the given diagram, we have:

$\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }$

So, x - 9 is positive when x → 9⁺. Therefore, |x -9) = x - 9

$\mathbf{ \lim_{x \to 9^+} (\dfrac{x-9}{-x^2-34+387}) }$

Simplifying the quadratic equation, we have:

$\mathbf{ \lim_{x \to 9^+} (-\dfrac{1}{x+43}) }$

Replacing the value of x = 9

$\mathbf{ = (-\dfrac{1}{9+43}) }$

$\mathbf{ = -\dfrac{1}{52} }$

$\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }$

-x²-63x+568 is positive when x → 8⁻.

Thus |-x²-63x+568| = -x²-63x+568

$\mathbf{ \lim_{x \to 8^-} (\dfrac{1}{x+71}) }$

$\mathbf{=\dfrac{1}{8+71} }$

$\mathbf{=\dfrac{1}{79} }$

$\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }$

x -7 is positive, therefore |x-7| = x - 7

$\mathbf{ \lim_{x \to 7^+} (\dfrac{-x^2-17x+168 }{x-7}) }$

$\mathbf{ \lim_{x \to 7^+} (-x-24)}$

$\mathbf{ \lim_{x \to 7^+} (-7-24)}$

= -31

$\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }$

x-6 is negative when x → 6⁻. Therefore, |x-6| = -x + 6

$\mathbf{ \lim_{x \to 6^-} (\dfrac{-x+6 }{-x^2-86x+552}) }$

$\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{x+92}) }$

$\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{6+92}) }$

$\mathbf{ =\dfrac{1}{98}}$

Learn more about calculating the limits of x here:

brainly.com/question/1444047

#SPJ1

7 0

2 years ago