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VladimirAG [237]
3 years ago
14

Help me on this one please

Mathematics
1 answer:
sashaice [31]3 years ago
6 0

Answer:

I believe the answer would be 3, and 7.

Step-by-step explanation:

Here's why to start off (x-3)(x-7)=0

What would you plug into x to make it 0, well x=3, and x=7, because 3-3=0, 7-7=0


Hope this helped!!!, :)

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Can the sum of two mixed numbers be equal to 2?
igomit [66]
No it cant because a mixed number is over one unless that mixed number is negative
3 0
3 years ago
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Which of the following is more than 0.35 but less than 0.41
Hoochie [10]
Its B because 3/10ths is a little bit more than 0.35, but way less than 0.41
6 0
3 years ago
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Need geometry help ASAP please!
tatiyna

Answer:

1. 121 π  unit²

2. 143°

3. 151 unit²

Step-by-step explanation:

1.

Area of a circle is given by the formula  A = πr²

where

A is the area,

r is the radius of the circle

From the given diagram, we can see that the radius is 11, hence the area will be:

A=\pi r^2\\A=\pi (11)^2\\A=121\pi

The answer is  121\pi units^2

2.

The unshaded secctor and the shaded sector equals the circle. We know that circle is 360°. The unshaded sector has an angle of 217°. So the shaded part will be 360 - 217 = 143°

The measure of the central angle of the shaded sector is 143°

3.

Area of a sector is given by the formula  A=\frac{\theta}{360}*\pi r^2

Where

\theta is the central angle of the sector (in our case it is 143°)

r is the radius (which is 11)

Plugging in all the info into the formula we have:

A=\frac{\theta}{360}*\pi r^2\\A=\frac{143}{360}*\pi (11)^2\\A=150.99

<em>rounding to the nearest whole number, it is </em>151 units^2

3 0
4 years ago
Plz need help. is very urgent.
Luda [366]

Answer:

Its blurry can you take it again

Step-by-step explanation:

5 0
3 years ago
Given a cylinder with a radius of 5cm and a height of 8 cm, find the volume of the cylinder. Use 3.14 for T.
Radda [10]

Answer: V=628cm³

Step-by-step explanation:

V=(3.14)r²h

V=(3.14)(5²)(8)

V=628cm²

Hope this helps! Have a great night!

6 0
3 years ago
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