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4 years ago

Please help. Don’t understand this math problem!

Mathematics

1 answer:

Step2247 [10]4 years ago

3 0

Answer:

y=-3log(3)(x+3)

Step-by-step explanation:

because the vertical asymptote at x=-3, the graph shifted to the left 3 units from x=0

y=A.log base(b)(x+3)

the graph passes through 2 points (-2,0) and (1,-3)

A=-2, b=0 and see if it fit point(0,-3)

-3 =-2log(0)(0+3) it is undefined check both sides if equal / not equal

check point (1,-3) ( shifted left 3 points)

A=-3, b=3

-3=-3log(3)(3) it is right

y=-3log(3)(x+3)

i hope it is right

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A. Find the common factors of 12 and 20.

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a. Common factors of 12 and 20 = 2, 2

2. Highest common factor of 12 and 20 = 4

3 packs of workbooks and 5 packs of text books to be purchased.

Step-by-step explanation:

a.) To find the common factors of 12 and 20:

First of all, let us find the factors of 12 and 20

12 = 2 $\times$ 2 $\times$ 3

20 = 2 $\times$ 2 $\times$ 5

The common factors are underlined above i.e. 2, 2.

2. The highest common factor between 12 and 20 is 2 $\times$ 2 = 4

-------------------------------------------

Workbooks are in packs of 25

Textbooks are in pack of 15

The school needs exactly same number of workbooks and textbooks.

And we have to find such least number.

This least number can simply be calculated by finding the LCM (Least Common Multiple).

LCM of two numbers p and q is the minimum number which is completely divisible by both the numbers p and q.

Let us find LCM of 25 and 15 using factorization method:

15 = 3 $\times$ 5

25 = 5 $\times$ 5

The common part is considered only once:

So LCM = 3 $\times$ 5 $\times$ 5 = 75

Hence, number of packs of workbooks to be bought = $\frac{75}{25} = 3$

number of packs of textbooks to be bought = $\frac{75}{15} = 5$

So, the answer are:

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2. Highest common factor of 12 and 20 = 4

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4 years ago