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2 years ago

7

Pam likes to practice dancing while preparing for a math tournament. She spends 80 minutes every day practicing dance and math.

To help her concentrate better, she dances for 20 minutes longer than she works on math.
Write a pair of linear equations to show the relationship between the number of minutes Pam practices math every day (x) and the number of minutes she dances every day (y). (5 points)

1 answer:

RSB [31]2 years ago

7 0

Answer: 628

Step-by-step explanation:

I found it online

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Joe has collections of 35 DVD movies. He received eight of them as gifts. Jobot the rest of his movies over three years. If he b

Yuliya22 [10]

Hello!

Let's subtract the eight DVDs Joe received as gifts, because he didn't buy them on his own.

35 - 8 = 27

Now, divide the remaining CDs by the 3 years he collected his DVDs in.

27 ÷ 3 = 9

A N S W E R:

Joe bought 9 DVDs last year.

Good day!

8 0

3 years ago

M is the incenter of triangle ABC. Find the lengths of MD and DC. (Look at pic)

Deffense [45]

Answer:

Part 1) $MD=9\ units$

Part 2) $DC=12\ units$

Step-by-step explanation:

Step 1

Find the length of MD

we know that

The incenter is the intersection of the angle bisectors of the three vertices of the triangle. Is the point forming the origin of a circle inscribed inside the triangle

so

In this problem

$MD=ME=MF$ ------> is the radius of a circle inscribed inside the triangle

we have that

$MF=9\ units$

therefore

$ME=9\ units$

$MD=9\ units$

Step 2

Find the length of DC

we know that

In the right triangle MDC

Applying the Pythagoras theorem

$MC^{2} =MD^{2}+DC^{2}$

we have

$MD=9\ units$

$MC=15\ units$

substitute

$15^{2} =9^{2}+DC^{2}$

$DC^{2}=225-81$

$DC=\sqrt{144}\ units$

$DC=12\ units$

6 0

3 years ago

is it possible to complete the mapping diagram for Diagram C so it represents a function? If so, complete the diagram to show a

GaryK [48]

Yes because you have to complete diagram A and B then you can do C so it’s depends on which numbers if it’s function or not.

8 0

2 years ago

Solve the following problems:

siniylev [52]

Answer:

DB = 13 cm

Step-by-step explanation:

ΔCAB ≅ ΔDBA by ASA, so CA ≅ DB by CPCTC.

CA = 13 cm, so DB = 13 cm.

8 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