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ra1l [238]
3 years ago
8

When a number is a multiple of 6 what are possible values of the ones digit

Mathematics
1 answer:
umka21 [38]3 years ago
8 0
6,2,8,4,and 0 are the only possible value of the ones digit, because 
6*1=6, last digit is 6
6*2=12, last digit is 2
6*3=18, last digit is 8
6*4=24, last digit is 4
6*5=30, last digit is 0
6*6=36, last digit is 6
and the whole cycle goes over again.
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Which equation has a slope of -2 and passes through the point (1,-6)?
Naily [24]

Answer:

y=-2x-4

Step-by-step explanation:

Given that,

Slope, m = -2

It passes through the point (1,-6).

We need to find the equation of line that passes throgh the given point. It can be calculated as:

y-y_1=m(x-x_1)

Put y₁ = -6, x₁ = 1 and m = -2

y-(-6)=(-2)(x-1)\\\\y+6=-2x+2\\\\y+2x+6-2=0\\\\y=-2x-4

Hence, the equation of line is y=-2x-4.

7 0
3 years ago
1/6 + 3/4 = <br> 1 1/2 + 2/5= <br> ez points<br><br> ​
Lorico [155]

Answer:

your clapped

Step-by-step explanation:

4 0
2 years ago
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Paula caught a tarpon with a weight that was 10 times as great as the weight of a permit fish she caught . The total weight of t
olya-2409 [2.1K]
X equals weight of permit fish
x + 10X = 132
11x = 132
x= 12

permit fish= 12
tarpon=120

7 0
3 years ago
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Write an equation for the cubic polynomial function whose graph has zeroes at 2, 3, and 5.
TiliK225 [7]

we know that

For a polynomial, if x=a is a zero of the function, then (x−a) is a factor of the function. The term multiplicity, refers to the number of times that its associated factor appears in the polynomial.

So

In this problem

If the cubic polynomial function has zeroes at 2, 3, and 5

then

the factors are

(x-2)\\ (x-3)\\ (x-5)

Part a) Can any of the roots have multiplicity?

The answer is No

If a cubic polynomial function has three different zeroes

then

the multiplicity of each factor is one

For instance, the cubic polynomial function has the zeroes

x=2\\ x=3\\ x=5

each occurring once.

Part b) How can you find a function that has these roots?

To find the cubic polynomial function multiply the factors and equate to zero

so

(x-2)*(x-3)*(x-5)=0\\ (x^{2} -3x-2x+6)*(x-5)=0\\ (x^{2} -5x+6)*(x-5)=0\\ x^{3} -5x^{2} -5x^{2} +25x+6x-30=0\\ x^{3}-10x^{2} +31x-30=0

therefore

the answer Part b) is

the cubic polynomial function is equal to

x^{3}-10x^{2} +31x-30=0

7 0
3 years ago
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Hurry please I’m running out of time! Show work
agasfer [191]

Answer:125

Step-by-step explanation:

x*25=125 x=125

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