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Anon25 [30]
2 years ago
5

PLEASE HELP!! WILL MARK BRAINLIEST IF RIGHT!!!

Mathematics
1 answer:
denpristay [2]2 years ago
4 0

Answer:

images are:

W'(-5,0)

X'(0,-9)

Y'(-9,-6)

Z'(-6,-2)

Step-by-step explanation:

use formula p(x,y)=p'(y,-x)

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Place 145 in the place value chart
Alexandra [31]

Answer: The 1 goes in the hundreds place and the 4 goes in the tens place and the 5 goes in the ones place

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
I will mark brainliest IF CORRECT
umka2103 [35]

Answer:

1 - (π/6) - (√3/4)

Step-by-step explanation:

First, calculate the area of ABCD. That's just 1*1 = 1 un^{2}

Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.

Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.

Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = \frac{1}{2} AD = \frac{1}{2}. MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.

Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.

The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.

The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.

Hope this helps, have a great day.

3 0
2 years ago
I need help..........
Marat540 [252]
Answer: 3rd one
Rearrange the original equation so it fits the model of : ax^2+bx+c=0
Then use the quadratic formula to find all possible answers.
5 0
3 years ago
What is the value of the expression below?<br><br> 0.72/0.36*2
Tanya [424]

Answer:

<h2>1</h2>

Step-by-step explanation:

0.72/0.36 x 2

0.72/0.72

1

I'm always happy to help :)

8 0
3 years ago
I don't know how to solve it.
Nataly [62]

Step-by-step explanation:

Take the first derivative

\frac{d}{dx} ( {x}^{3}  - 3x)

3 {x}^{2}  - 3

Set the derivative equal to 0.

3 {x}^{2}  - 3 = 0

3 {x}^{2}  = 3

{x}^{2}  = 1

x = 1

or

x =  - 1

For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).

For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.

Since we are going to Positve to negative slope, we have a local max at x=-1

Plug in -1 for x into the original function

( - 1) {}^{3}  - 3(  - 1) = 2

So the local max is 2 and occurs at x=-1,

For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.

Since we are going to decreasing to increasing, we have minimum at x=1,

Plug in 1 for x into original function

{1}^{3}   - 3(1)

1 - 3 =  - 2

So the local min occurs at -2, at x=1

8 0
2 years ago
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