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densk [106]
3 years ago
5

Any one knows? Please help

Mathematics
2 answers:
ella [17]3 years ago
4 0
Hey there ur answer will be a. 9,12,5
valkas [14]3 years ago
3 0

Answer:

1. 12

2.

a can be a triangle (obtuse)

b can be a triangle (acute)

Step-by-step explanation:

You might be interested in
You can represent an odd integer whith the expression 2n+1, where n is any integer . Write and solve an equation to find three c
oksian1 [2.3K]
Any odd number can be expressed by 2n+1.

For example,  

2n+1=111
2n=110
n=110/2=55

means that 111 is 2n+1 for n=55


Thus if an odd number is 2a+1, the next few numbers are as follows:

2a+1, 2a+2, 2a+3, 2a+4, 2a+5

So 2a+1, 2a+3 and 2a+5 are 3 consecutive odd numbers.


Back to our problem: 

three consecutive odd numbers whose sum is 63 are:

(2n+1)+(2n+3)+(2n+5)=63

6n+9=63

6n=63-9=54

n=54/6=9


2n+1=2*9+1=18+1=19, the 2 next odd numbers are 21 and 23


Answer: 19, 21, 23



5 0
3 years ago
HELP DUE IN 10 MINUTES!! WILL GIVE BRAINLEST IF CORRECT!!!!
Harman [31]

Answer:

<u>Given equation:</u>

  • y = 10x - 32

<u>To find the y-intercept, evaluate the equation with x = 0:</u>

  • y = 10*0 - 32
  • y = 0 - 32
  • y = -32

<u>To find the x-intercept, evaluate the equation with y = 0:</u>

  • 0 = 10x - 32
  • 10x = 32
  • x = 32/10
  • x = 3.2
8 0
3 years ago
Under the best conditions, a sprout has a 40% probability of blooming. If two sprouts are selected at random, what is the probab
Elenna [48]

Answer:

16%

Step-by-step explanation:

Probability of sprout blooming = 40% = 40/100 = 2/5

Probability of two sprouts blooming = 2/5 × 2/ 5

= 4/25 = 16%

5 0
4 years ago
Read 2 more answers
The radius of a sphere is increasing at a rate of 5 mm/s. How fast is the volume increasing (in mm3/s) when the diameter is 40 m
jeka94

9514 1404 393

Answer:

  8000π mm^3/s ≈ 25,133 mm^3/s

Step-by-step explanation:

The rate of change of volume is found by differentiating the volume formula with respect to time.

  V = 4/3πr^3

  V' = 4πr^2·r'

For the given numbers, this is ...

  V' = 4π(20 mm)^2·(5 mm/s) = 8000π mm^3/s ≈ 25,133 mm^3/s

_____

<em>Additional comment</em>

By comparing the derivative to the area formula for a sphere, you see that the rate of change of volume is the product of the area and the rate of change of radius. This sort of relationship will be seen for a number of different shapes.

4 0
3 years ago
Please help me with question number 2
Mice21 [21]

Answer:

2. The area of the side walk is approximately 217 m²

3. The distance away from the sprinkler the water can spread is approximately 11 feet

4. The area of the rug is 49.6

Step-by-step explanation:

2. The dimensions of the flower bed and the sidewalks are;

The diameter of the flower bed = 20 meters

The width of the circular side walk, x = 3 meters

Therefore, the diameter of the outer edge of the side walk, D, is given as follows

D = d + 2·x (The width of the side walk is applied to both side of the circular diameter)

∴ D  = 20 + 2×3 = 26

The area of the side walk = The area of the sidewalk and the side walk = The area of the flower bed

∴ The area of the side walk, A = π·D²/4 - π·d²/4

∴ A = 3.14 × 26²/4 - 3.14 × 20²/4 = 216.66

By rounding to the nearest whole number, the area of the side walk, A ≈ 217 m²

3. Given that the area formed by the circular pattern, A = 379.94 ft.², we have;

Area of a circle = π·r²

∴ Where 'r' represents how far it can spread, we have;

π·r² = 379.94

r = √(379.94 ft.²/π) ≈ 10.997211 ft.

Therefore, the distance away from the sprinkler the water can spread, r ≈ 11 feet

4. The circumference of the rug = 24.8 meters

The circumference of a circle, C = 2·π·r

Where;

r = The radius of the circle

π = 3.1

∴ For the rug of radius 'r', C = 2·π·r = 24.8

r = 24.8/(2·π) = 12.4/π = 12.4/3.1 = 4

The area = π·r²

∴ The area of the rug = 3.1 × 4² = 49.6.

8 0
3 years ago
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