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3 years ago

7

Aria is buying snacks for a party. She will buy one pizza for $20 and some garlic rolls for $2 each. Write an expression to repr

esent the total money she spends buying snacks for the party.

1 answer:

sineoko [7]3 years ago

3 0

$20+$2w is the correct answer… w is just the unknown variable so you can change it to any letter you may wish! Hope this helped!

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A. What is the explicit definition for this sequence?<br><br>b. How far does she run on day 19?

creativ13 [48]

Answer:

(a) $a_n=1+\frac{1}{6}(n-1)$

(b) 4 miles

Step-by-step explanation:

Given:

The sequence of run is an arithmetic sequence.

Miles traveled on day 1 = 1 mile

Miles traveled on day 7 = 2 miles

(a).

Explicit formula for an arithmetic sequence is given as:

$a_n=a_1+(n-1)d\\a_n\to n^{th}\ term\\a_1\to first\ term\\n\to number\ of\ terms\\d\to \textrm{common difference}$

Here, $a_1=1,a_7=2$

So,

$a_7=a_1+(7-1)d\\2=1+6d\\6d=2-1\\d=\frac{1}{6}$

Therefore, the explicit definition of the above sequence is given as:

$a_n=1+\frac{1}{6}(n-1)$

Where, $a_n$ is the miles covered on $n^{th}$ day.

(b)

In order to find the distance covered by her on day 19, we plug in 19 for n in the above formula. This gives,

$a_{19}=1+\frac{1}{6}(19-1)\\\\a_{19}=1+\frac{1}{6}(18)\\\\a_{19}=1+3=4\ mi$

Therefore, the number of miles covered by her on day 19 is 4 miles.

3 0

3 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago

Can someone please solve

pentagon [3]

Answer:

78

Step-by-step explanation:

Evaluate 6 (2 x^2 - 5) where x = -3:

6 (2 x^2 - 5) = 6 (2×(-3)^2 - 5)

Hint: | Evaluate (-3)^2.

(-3)^2 = 9:

6 (2×9 - 5)

Hint: | Multiply 2 and 9 together.

2×9 = 18:

6 (18 - 5)

Hint: | Subtract 5 from 18.

| 1 | 8

- | | 5

| 1 | 3:

6×13

Hint: | Multiply 6 and 13 together.

6×13 = 78:

Answer: 78

7 0

2 years ago

Two trains are 500 miles apart when they first enter a collision course. If, after two hours, the distance between them is 300 m

MrRissso [65]

Answer:

The faster train goes 60mph

Step-by-step explanation:

s= speed

s+20 is the faster train.

Then you are given:

200/s+s+20=2

200=2*(2s+20)

4s+40=200

4s=160

s=40 is the slower train

40+20= speed of faster one

s=60 ;)

6 0

3 years ago

Find the area and the circumference of a circle with Diameter 6 yd . Use the value 3.14 for pi , and do not round your answers .

Maurinko [17]

Step-by-step explanation:

i hope this helps its kind of hard to see but i tried lol

4 0

3 years ago