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The factorised form of 6m + 18 is given by the equation: 6(m + 3).

Given :

Equation - 6m + 18

Solution :

Factorisation is the method of diminishing the bracket of a quadractic condition, rather than growing the bracket and changing over the condition to a item of variables which cannot be decreased assist.

To factorise the equation 6m + 18 we have to follow some steps:

In first step we have to write 18 = [tex]6\times 3[/tex] so the equation becomes:

[tex]\rm = 6m + (6\times3)[/tex]  ----- (1)

In second step we have to take 6 common from equation (1):

[tex]\rm=6(m+3)[/tex]

So the factorised form of 6m + 18 is given by the equation: 6(m + 3)

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We want to answer different things about an exponential decay, the answers are:

So we know that the original price of the computer is $1250 and the value decays by 10% each year.

So in year 1, the new value of the computer will be:

V = $1250*(1 - 0.1) = $1250*0.9

On year 2, the new value will be:

V = ( $1250*0.9)*0.9 = $1250*(0.9)^2

And so on.

a) This sequence is a geometric sequence because each term is a constant times the previous term, where the constant is 0.9.

[tex]A_1 = 1250\\A_2 = 0.9*1250 = 1125\\A_3 = 0.9^2*1250 = 0.9*1125 = 1012.5 \\...[/tex]

b) The explicit formula for a geometric sequence is:

[tex]A_n = k^{n-1}*A_1[/tex]

where:

k is the constant, in this case, is 0.9

A1 is the first term of the sequence, in this case, is 1250

Then we have:

[tex]A_n = (0.9)^{n-1}*1250[/tex]

c) Here we just need to replace n by 6 in the above formula:

[tex]A_6 = (0.9)^5*1250 = 738.1125[/tex]

This means that the price of the computer in the 6th year is $738.10

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Final answer:

The series 5+25+125+625+3125+15625 can be rewritten in sigma notation as Σ 5*5^(n-1) from n=1 to 6, representing a geometric series with a common ratio of 5 and 6 terms.

Explanation:

To rewrite the series 5+25+125+625+3125+15625 using sigma notation, we first need to identify the pattern of the series. We notice that each term is 5 times the previous term, which is a characteristic of a geometric series. A geometric series can be expressed in sigma notation as Σ a*r^(n-1) from n=1 to N, where a is the first term, r is the common ratio between terms, and N is the number of terms.

In this series, a = 5, and r = 5. The series has 6 terms, so N = 6. Therefore, the series in sigma notation is Σ 5*5^(n-1) from n=1 to 6.

Answer:

His rate of discount=10%

Step-by-step explanation:

Step 1: Get the discount amount

Discount amount=List price-sale price

where;

List price=$800

Sale price=$720

replacing;

Discount amount=(800-720)=80

Discount amount=$80

Step 2: Calculate discount rate

The discount rate can be calculated as;

discount rate=(discount/List price)×100

where;

discount=$80

List price=$800

replacing;

discount rate=(80/800)×100=10%

discount rate=10%

To find a counterfeit coin among 8 coins with a balance scale, divide the coins and use up to 3 weighings. The algorithm involves grouping and comparing the weights, identifying the counterfeit in the imbalanced group.

Algorithm to Find a Counterfeit Coin among 8 Coins

To find a counterfeit coin among 8 coins with a minimum number of weighings on a balance scale, we need at most 3 weighings. Here is the step-by-step algorithm:

With this algorithm, the maximum number of weighings needed to determine the counterfeit coin is three.

The value of w is 14.

The image given is a right angled triangle. In order to determine the value of the hypotenuse, w, tan would be used.

Cos 30 = adjacent / hypotenuse

√3 / 2 = 7√3 /  hypotenuse

hypotenuse = 7√3 ÷ √3/2  = 14

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A proportion is an equation that shows two ratios are equivalent, such as 1/2 = 3/6. It is commonly used with unit rates and unit scales in various fields including science, engineering, and everyday measurements.

An equation stating that two ratios are equivalent is known as a proportion. For example, the equation 1/2 = 3/6 is a proportion because it shows that the two ratios have the same value. Proportions are very useful in various aspects of mathematics and science. Notably, they are essential when dealing with unit rates and unit scales, where comparisons between two measurements are made.

A unit rate is a special kind of ratio where one of the quantities is one, such as 55 miles per hour (55/1 mph). It describes how many units of the first type of measurement corresponds to one unit of the second type of measurement. Similar in concept, a unit scale compares the dimensions of an actual object to a model or drawing. For instance, a map might have a unit scale that reads 1 inch = 100 feet, which would be written as 1 inch/100 feet.

When applying proportions in real-world scenarios, we can create conversion factors to switch from one measurement to another. This is directly applicable to cooking recipes, chemistry with mole ratios, or engineering with the transformer equation, where ratios of different quantities are kept consistent.

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Answer:

d. $10,234.80

Step-by-step explanation:

we are given

The equipment will cost $26,000

so, Amount is $26000

A=26000

should be invested today at 12%

r=12%=0.12

It is compounded semiannually

so, n=2

t=8

now, we can use formula

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]

now, we can plug values

[tex]26000=P(1+\frac{0.12}{2})^{2\times 8}[/tex]

now, we can solve for P

[tex]P=10234.80338[/tex]

Answer : The equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

Explanation :

Half-life = 5.27 years

Formula used :

[tex]a=\frac{a_o}{2^n}[/tex]        ............(1)

where,

a = amount of reactant left after n-half lives

[tex]a_o[/tex] = Initial amount of the reactant

n = number of half lives

And as we know that,

[tex]n=\frac{t}{t_{1/2}}[/tex]        ..........(2)

where,

t = time

[tex]t_{1/2}[/tex] = half-life  = 5.27 years

Now equating the value of 'n' from (2) to (1), we get:

[tex]a=\frac{a_o}{2^{(\frac{t}{t_{1/2}})}}[/tex]       ...........(3)

[tex]a=\frac{a_o}{2^{(\frac{t}{5.27})}}[/tex]

[tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

Therefore, the equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)^(t/h), where t is the number of years and h is the half-life of the isotope.

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)(t/h), where t is the number of years and h is the half-life of the isotope.

For example, for the radioisotope cobalt-60 with a half-life of 5.27 years, you can use the equation Percent remaining = (100) x (1/2)(t/5.27) to determine the percent remaining after t years.

Let's say you want to find the percent remaining after 10 years, you would substitute t with 10 in the equation, like this: Percent remaining = (100) x (1/2)(10/5.27). Simplify the equation to find the answer.

Answer:

the answer is p=40h, hope this helps!!!

Step-by-step explanation:

                     Option: A is the correct answer.

                         [tex]x=-2(y+3)^2-1[/tex]

We know that any general equation of a right or left open parabola is given by:

                           x=a(y-k)^2+h

where if a<0 then the parabola open to the left and if a>0 then the parabola opens right.

Also the vertex of the parabola is: (h,k)

Now here we are given:

Vertex (h,k)=(-1,-3)

Also a= -2 in each of the options.

Hence, we have the equation of parabola as:

                [tex]x=-2(y-(-3))^2+(-1)\\\\x=-2(y+3)^2-1[/tex]

Hence, the equation of the parabola is:

                  [tex]x=-2(y+3)^2-1[/tex]