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Anvisha [2.4K]
3 years ago
14

Graph the equation y=-x^2+12x-32y on the accompanying set of axes. You must plot 5 points including the roots and the vertex.

Mathematics
1 answer:
Hitman42 [59]3 years ago
6 0
Use desmos ... that’s what i use for math
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How do you write 21% as a decimal
Goshia [24]

Answer:

0.21

Step-by-step explanation:

if you write it as a fraction it would be 21/100

and if that is made to a decimal it would be 0.21 .

8 0
3 years ago
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The vertices of a rectangle are (1, -3), (4,0), (0,4) and (-3,1). What is the area of the rectangle? ​
motikmotik

Answer:

48

Step-by-step explanation:

You can use a graphing calculator like desmos or draw your own graph to figure this out. Start by graphing each of those points. Then count the width and height. Multiply them to get the area. The width and height of the rectangle were 6 and 8, so the area is 48

3 0
3 years ago
Explain how you can find the volume of the figure below. Then, solve- what is the volume of the figure? Make sure to include you
mote1985 [20]

Answer: a) The figure can be reasonably divided into two geometries:

• a rectangular prism

• a hemisphere.

b) The volume of the rectangular prism is given by

  V = lwh

  V = (10 cm)(5 cm)(4 cm) = 200 cm³

The volume of the hemisphere is given by

  V = (2π/3)r³

  V = (2π/3)(3 cm)³ = 18π cm³

c) The total volume of the figure is

  total volume = (prism volume) + (hemisphere volume)

  V = 200 cm³ + 18π cm³

  V ≈ 256.549 cm³

Step-by-step explanation:

6 0
3 years ago
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mark mowed 1/3 of a football field. he mowed equal amounts over 2 hours how much of a field did mark mow each hour?
grigory [225]

Answer: 1/6


Step-by-step explanation: ok, so he had mowed 1/3 of a field in 2 hours. Take 1/3 and divide by 2 hours to get 1/6


4 0
3 years ago
Help me pleaseee.....
Lana71 [14]

Answer:

titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

1−x

2

∣=

2

(2x

2

−1)\Leftright∣cos\alp+sin\alp∣=

2

(2cos

2

\alp−1)

\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

2

cos(\alp−

4

π

)∣=N

2

cos(2\alp)\Right\alp∈[0;

4

π

]∪[

4

3π

;π]

1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

4

π

]

\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−

4

π

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

4

3π

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

4

π

)=cos(2\alp)…

1

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Display

6 0
4 years ago
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