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3 years ago

you buy two rolls of wrapping paper. each roll has the same lateral surface area. what is the diameter of roll B?

Mathematics

1 answer:

scoundrel [369]3 years ago

8 0

The area of lateral surface is equal that the perimeter multiply by the length. So the lateral surface area of roll A is 3*π*30=90π. Roll B: d*π*24=24dπ. Put them together: 90π=24dπ. So d=3.75 in.

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Write the equation of the line in slope-intercept form that has: slope: -1 and point: (1, 1)

lara31 [8.8K]

Answer:

f(x) = -1x + 1

Step-by-step explanation:

f(x) = mx + b

the equation above is slope intercept form. in order to make it you need a slope and a y intercept. you were given the slope, so now you need a y intercept. but where would you find that??? look no farther than the point you were given, (1,1). the y intercept is the second number in the parentheses. next you take your slope and your y-intercept and place them in.

f(x) = -1x + 1

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3 0

2 years ago

Plz help ASAP

anygoal [31]

Answer:

B.40

Step-by-step explanation:

4 0

2 years ago

Round the answer to the nearest

kykrilka [37]

By knowing the blood pressure and applying the quadratic formula, the age of a man whose normal blood pressure is 129 mm Hg is 40 years old.

<h3>How to use quadratic equations to determine the age of a man in terms of blood pressure</h3>

In this problem we have a quadratic function that models the blood pressure as a function of age. As the latter is known, we must use the quadratic formula to determine the former:

129 = 0.006 · A² - 0.02 ·A + 120

0.006 · A² - 0.02 · A - 9 = 0

$A = \frac{0.02 \pm \sqrt{0.006^{2}-4\cdot (0.006)\cdot (- 9)}}{2\cdot (0.006)}$

A = 1.667 + 38.733

A = 40

By knowing the blood pressure and applying the quadratic formula, the age of a man whose normal blood pressure is 129 mm Hg is 40 years old.

To learn more on quadratic equations: brainly.com/question/1863222

#SPJ1

4 0

1 year ago

Please solve 1 thru 4

jolli1 [7]

Answer:

1.triangular prism

2.two dimensional

3.faces

4. surface area

5 0

2 years ago

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Divide the combinations. 60C3 / 15C3

Mariulka [41]

I hope you know what '!' means
2!=2
3!=6
4!=24
basicalkly times every natural number including and before that number

nCr= $\frac{n!}{r!(n-r)!}$
so
$\frac{60C3}{15C3}$ =
$\frac{ \frac{60!}{3!(60-3)!} }{ \frac{15!}{3!(15-3)!} }$ =
$( \frac{60!}{3!(60-3)!})(\frac{3!(15-3)!}{15!})$ =
$( \frac{60!}{3!(57)!})(\frac{3!(12)!}{15!})$ =
$\frac{(60!)(3!)(12!)}{(3!)(57!)(15!)}$ =
$\frac{(60!)(12!)}{(57!)(15!)}$ =
$\frac{(60!)}{(57!)(15)(14)(13)}$ =
$\frac{(60)(59)(58)}{(15)(14)(13)}$ =
$\frac{(30)(59)(58)}{(15)(7)(13)}$ =
$\frac{(6)(59)(58)}{(3)(7)(13)}$ =
$\frac{6844}{91}$

4 0

3 years ago

Read 2 more answers