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2 years ago

Diego made the shape on the left and Elena made the shape on the right. Each shape uses 5 circles.

Mathematics

1 answer:

ruslelena [56]2 years ago

4 0

Answer:

Step-by-step explanation:

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Wayne is going to paint the side of the house shown in the diagram. What is the area that will be painted? Please explain how u

goblinko [34]

First find the area of the rectangle by multiplying the length and width of the rectangle, so 12 x25=300. Then find the area of the triangle, which is length x height/ 2. So 8 x 25=200/2=100. Then add the two answers together to get 400.

3 0

3 years ago

For tax purposes, a car rental company assumes each car in their fleet depreciates by 6% per year. If the initial value of a car

n200080 [17]

The value of a car after 10 years be $20898.27 if the car rental company assumes each car in their fleet depreciates by 6% per year option (a) $20898.27 is correct.

<h3>What is an exponential function?</h3>

It is defined as the function that rapidly increases and the value of the exponential function is always a positive. It denotes with exponent $\rm y = a^x$

where a is a constant and a>1

We can solve this problem by exponential function:

The word depreciate means the price is decreasing.

We can find the value be when the car is 10 years old:

p = 38800(1 - 0.06)¹⁰

p = 38800(0.94)¹⁰

p = 20898.266 ≈ $20898.27

Thus, the value of a car after 10 years be $20898.27 if the car rental company assumes each car in their fleet depreciates by 6% per year option (a) $20898.27 is correct.

Learn more about the exponential function here:

brainly.com/question/11487261

#SPJ1

6 0

1 year ago

32x - 4 = 4x2 + 60 For the equation shown, choose the description of the solutions.

Furkat [3]

All to one side:

4x^2-32x+64 = 0,

divide by 4:

x^2 -8x + 16 = 0

apply formula or see that it is (x-4)^2!, so only x =4 is a root and it is double.

3 0

3 years ago

Read 2 more answers

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago

PLEASE PLEASE PLEASE HELP ME. IM STUCK IN A SUMMER SCHOOL AND IF I DONT GET GOOD GRADES ILL GET IN BIG TROUBLE AND IM TOTALLY LO

Mrrafil [7]

Answer:

We know that in the box there are:

4 twix

3 kit-kat

Then the total number of candy in the box is:

4 +3 = 7

Here we want to find the probability that we draw two twix.

All the candy has the same probability of being drawn from the box.

So, the probability of getting a twix in the first drawn, is equal to the quotient between the number of twix and the total number of candy in the box, this is:

p = 4/7

Now for the second draw, we do the same, but because we have already drawn one twix before, now the number of twix in the box is 3, and the total number of candy in the box is 6.

this time the probability is:

q = 3/6 = 1/2

The joint probability is the product of the individual probabilities, so here we have

P = p*q = (4/7)*(1/2) = 2/7

b) same reasoning than in the previous case:

For the first bar, the probability is:

p = 3/7

for the second bar, the probability is:

q = 2/6 = 1/3

The joint probability is:

P = p*q = (3/7)*(1/3) = 1/7

c) Suppose that first we draw a twix.

The probability we already know that is:

p = 4/7

Now we want another type, so we need to draw a kit-kat, the probability will be equal to the quotient between the remaining kit-kat bars (3) and the total number of candy in the box (6)

q = 3/6

The joint probability is:

P = p*q = (4/7)*(3/6) = 2/7

But, we also have the case where we first draw a kit-kat and after a twix, so we have a permutation of two, then the probability in this case is:

Probability = 2*P = 2*2/7 = 4/7

3 0

3 years ago