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3 years ago

Which of the following is the mathematical representation of Boyle's law?

1 answer:

3 0

The mathematical representation of Boyle's law is P1V1 = P2V2. The answer to your question is C. I hope that this is the answer that you were looking for and it has helped you.

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The graph shows the relationship between the number of pounds of pears purchase and the total cost of the pears.

Katarina [22]

The answer is $1.75 because that’s what the graph shows per pound of pears

6 0

2 years ago

A blank is a unit rate with a denominator of 1 unit

Dima020 [189]

It is 3. That's the correct answer

6 0

3 years ago

Frank is making a pennant in the shape of a triangle for his senior class photo. He wants the base length of this triangle to be

Sav [38]

Answer:

Height should be ≤ 9 inches.

Step-by-step explanation:

Given:

Frank is making a pennant in the shape of a triangle for his senior class photo.

Base of triangle = 6 in

Area of triangle ≤ 27 $in^2$

Let height of the triangle be h

Now we now that,

Area of triangle = $\frac{1}{2}\times base \times height$

$\frac{1}{2}\times 6 \times h \leq 27\\\\3h\leq 27\\\\h\leq \frac{27}{3}\\\\h\leq 9 \ in.$

Hence the height of the triangle must be at most or ≤ 9 inches.

5 0

2 years ago

What is the fourth term in the binomial expansion of (2x – y) 7?

nata0808 [166]

Answer:

$(7C4) (2x)^4 (-y)^{7-4}$

And replacing we got:

$35 (2^4) x^4 (-y)^{-3}$

And then the final term would be:

$-560 x^4 y^3$

Step-by-step explanation:

For this case we have the following expression:

$(2x-y)^7$

And we can use the binomial theorem given by:

$(x+y)^n =\sum_{k=0}^n (nCk) x^k y^{n-k}$

And for this case we want to find the fourth term and using the formula we have:

$(7C4) (2x)^4 (-y)^{7-4}$

And replacing we got:

$35 (2^4) x^4 (-y)^{-3}$

And then the final term would be:

$-560 x^4 y^3$

8 0

2 years ago

You need to create a 4-character password that consists of four English letters and/or digits if repetition is allowed. How ma

Nataly_w [17]

123a
123q
123w
123e
123r
123t
123y
123u
123i
123o
123p
123s
123d
123f
123g
123h
123j
123k

I thought this would be easy lol, I tried, sorry for not getting an answer

3 0

3 years ago