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3 years ago

Factor x2+12x+35 , I appreciate all the kind generous help I can receive

Mathematics

1 answer:

stellarik [79]3 years ago

8 0

So, to factor out the equation x²+12x+35, first multiply the first and last numbers.
1×35=35
So, you need to find the multiples of 35 that add up to 12.
So, you find 7 and 5. These two numbers are the magic numbers that will be used as the correct factored form of x²+12x+35, because:
>7×5=35
>7+5=12
So, using these two numbers, you get your factored form as:
(x+5)(x+7)

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In short, Your Final Answer would be: 730

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In Figure 2, Red boxes represent "Chocolate flavor ice-cream" which is 6 in number whereas Blue boxes represent "Vanila".

1 box contains = 1440/15 = 96

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In short, Your Final Answer would be: 576

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In Figure 3, Red boxes represent "diet sodas" which is 5 in number whereas Blue boxes represent "Regular sodas".

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4 years ago

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Q # 15 please somebody help me

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4 years ago

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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

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4 years ago

What is the approximate area of the figure? Help pls.....

Vaselesa [24]

Okay, first we need to calculate the area of the rectangle.

Area of Rectangle- 16 x 6 = 96

Next, we need to calculate the area of the semicircles.

Formula: Radius x Radius x pi (3.14) = Area of circle

Since we know that the diameter is 6 we can figure out the radius.
The radius is half of the diameter. So our radius is 3.

3 x 3 x 3.14 = 28.26

Since it's a semi circle we now need to divide by 2.

28.26 divided by 2 = 14.13

Then, we add.

Area of rectangle
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96 + 14.13 + 14.13 = Total Area of The Figure

So our answer is 124.26

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Step-by-step explanation:

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