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2 years ago

Only questions 5 and 6 pls.

Mathematics

1 answer:

lesantik [10]2 years ago

4 0

theres no pic so how do we answer

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What's the answer..............

ruslelena [56]

Aye the Answer is D ~hope that helps :)

7 0

3 years ago

Amy needs to mail a gift card to a friend. She uses 47-cent stamps and 6-cent stamps to pay $2.42 in postage. How many of each s

AVprozaik [17]

Answer:

Answer:Amy used 4 41-cent stamps and 8 6-cent stamps.

Step-by-step explanation:

Let x represent the number of 41-cent stamps that Amy used. Let y represent the number of 6-cent stamps that Amy used.

41 cents = 41/100 = $0.41

6 cents = 6/100 = $0.06

She uses 41-cent stamps and 6-cent stamps to pay $2.12 in postage. It means that

0.41x + 0.06y = 2.12

Multiplying through by 100, it becomes

41x + 6y = 212

6y = 212 -41x

We would test for corresponding values of x and y that satisfies the equation and they must be whole numbers.

If x = 3,

6y = 212 - 41 × 3 = 89

y = 89/6 = 14.8333

If x = 4,

6y = 212 - 41 × 4 = 48

y = 48/6 = 8

Answer:Amy used 4 41-cent stamps and 8 6-cent stamps.

Step-by-step explanation:

Let x represent the number of 41-cent stamps that Amy used. Let y represent the number of 6-cent stamps that Amy used.

41 cents = 41/100 = $0.41

6 cents = 6/100 = $0.06

She uses 41-cent stamps and 6-cent stamps to pay $2.12 in postage. It means that

0.41x + 0.06y = 2.12

Multiplying through by 100, it becomes

41x + 6y = 212

6y = 212 -41x

We would test for corresponding values of x and y that satisfies the equation and they must be whole numbers.

If x = 3,

6y = 212 - 41 × 3 = 89

y = 89/6 = 14.8333

If x = 4,

6y = 212 - 41 × 4 = 48

y = 48/6 = 8

8 0

3 years ago

Read 2 more answers

How many modes does the data set shown below contain?<br><br><br> 1, 3, 3, 4, 6, 6, 7, 7, 7, 10, 12

kipiarov [429]

Answer:

There are 3 modes

Step-by-step explanation:

There are two 3s, two 6s, and three 7s

4 0

3 years ago

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

In the model below what is x and how do I solve it

tiny-mole [99]

Answer:so that’s a interior triangle so the mean your 2 points in the inside and the equation on the outside.

X=6

Step-by-step explanation:

(8x-4)+(3x+17)=(17x-23)

Combined our alike terms on the left side of the equation. So 8x+3x and -4+17.

11x+13=17x-23

Now this is what the equation should look like after combining ur alike terms.

11x+13=17x-23

-11x. -11x.

now u take the lowest x which is 11x and mince that from 17x but u only mince if it’s a positive but is it’s a negative u add it.

13=6x-23 so if u moved the x on the right side u move non variable number to the left but if u moved the x to the left u move the non variable number to the right so in this case the non variable number would be -23 so now u add 23 on both sides.

36=6x and now u just divide the 36 by 6 and remember to always divide by the x

Now the answer is x=6

7 0

3 years ago