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professor190 [17]
3 years ago
14

What are common mutipiles of 5 and 8

Mathematics
1 answer:
Natalija [7]3 years ago
5 0

Answer:List the multiples of 8 and stop when you find a multiple of both 5 and 6. Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, … Stop at 120 as it is a multiple of both 5 and 6. So, the LCM of 5, 6 and 8 is 120

Step-by-step explanation:

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In how many different ways can 6 runners be assigned to 6 lanes at the start of a race?
Vedmedyk [2.9K]
Well you have runner 1 in lane 1. runner 1 in lane 2 and so on. so it becomes 1,1 1,2 1,3 1,4 1,5 1,6 and then 2,1 2,2 2,3 2,4 2,5 2,6 then 3, and so on
5 0
4 years ago
What are the amplitude, period, and midline of f(x)=-4 cos (2x-n)+3?
tangare [24]
To find the amplitude and period you need to be familiar with the following equation. Also you need to know that the standard cos has a period of 2\pi and the midline is a line that runs between the max and min of the y-values of the function.

Equations:
f(x) = A cos(Bx +C) + D
f(x) = -4 cos(2x -n) + 3

A = amplitude = |-4| = 4
B = 2
C = phase shift = n = 0 
D = vertical shift = midline = 3

Amplitude = 4

Find the period:
Period = \frac{2\pi}{B} = \frac{2\pi}{2} = \pi

Find the midline:
We know that the amplitude is 4  so we have a range from -4 to 4. The standard y = cos(x) has its midline at 0 so y = 0. This is also true for y = -4 cos(x). In your equation though, you have a vertical shift of +3 so this changes our midline. With an amplitude of 4, which gives us a range from -4 to 4(our y-values), the shift moves this up by 3 so that means we will have new y-values and a range of -1 to 7. Now we need to find the midline(the middle of our y-values) of our new range. We can find this by using the following formula
Midline = \frac{Y_1 + Y_2}{2} = \frac{-1+ 7}{2} = \frac{6}{2} = 3

Midline:
y = 3

Note, in the following equations that D = 3 = midline

y = A (Bx+C) + D
y = -4 (2x + n) + 3

Also, the picture that is attached is what your equation looks like when graphed.


7 0
3 years ago
-8x + 3y = 15<br> Calculate slope from an equation
ludmilkaskok [199]

Answer: 8/3

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Triangles A B C and H G L are congruent. Angles A B C and H G L are right angles. The length of hypotenuse A C is 15 and the len
Lisa [10]

Answer:

x = 4

Therefore, for the triangles to be congruent by HL, the value of x must be 4.

Step-by-step explanation:

Given: ΔABC and ΔHGL are congruent. ∠ABC = ∠HGL = 90°.

Length of hypotenuse AC = 15

Length of hypotenuse HL = 3x + 3

Length of AB = 9, Length of BC = 12 and Length of GL = 2x + 1.

Sol: ∵ ΔABC ≅ ΔHGL

Length of HL = Length of AC (corresponding parts of congruent triangles)

3x + 3 = 15

3x = 15 - 3

3x = 12

x = 12/3 = 4

Therefore, for the triangles to be congruent by HL, the value of x must be 4.

5 0
3 years ago
What is the solution to this equation 2x-x+9+3x-2=16
Alexxx [7]
First add liile terms so you equation will be 4x+7=16, then subtract 7 from both sides and you get 4x=9, finally divide both sides by 4 and you will get x=2.25
5 0
3 years ago
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