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2 years ago

A jar lid has a diameter of 42 millimeters.What is the circumference of the lid?

2 answers:

7 0

$\sf{Circumference = \pi \times diameter$

We are given that the diameter = 42 millimeters

$C = \pi \times 42\\\\\boxed{\bf{C = 42\pi \approx 131.95~mm}}$

The circumference of the lid is approximately 131.95 millimeters.

atroni [7]2 years ago

3 0

Circumference of a circle = π x diameter

= π x 42
= <span>131.946891451
= 132 mm</span>

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The ideal temperature of a freezer to store a particular brand of ice cream is 0°F, with a fluctuation of no more than 2°F. Whic

yan [13]

Answer:

| x - 0 | ≤ 2

Step-by-step explanation:

Given,

The ideal temperature of the freezer = 0° F,

Also, it can fluctuate by 2° F,

Thus, if x represents the temperature of the freezer,

Then, there can be two cases,

Case 1 : x > 0,

⇒ x - 0 ≤ 2

Case 2 : If x < 0,

⇒ 0 - x ≤ 2,

⇒ -( x - 0 ) ≤ 2,

By combining the inequalities,

We get,

| x - 0 | ≤ 2,

Which is the required inequality.

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3 years ago

You owe $300 on your credit card. For every month that you fail to make a payment, your balance increases by 1%. Find a formula

Vlada [557]

Answer:

$B(t) = 300(1.01)^{t}$

Step-by-step explanation:

The balance owed after t months can be given by an exponential function in the following format:

$B(t) = B(0)(1+r)^{t}$

In which B(0) is the initial balance owed and r is the growth rate, as a decimal.

You owe $300 on your credit card. For every month that you fail to make a payment, your balance increases by 1%.

This means, respectively, that $B(0) = 300, r = 0.01$

$B(t) = 300(1+0.01)^{t}$

$B(t) = 300(1.01)^{t}$

4 0

3 years ago

What is the correct answer for the question

polet [3.4K]

Answer:

2 $\frac{1}{12}$

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the sum of the distances of R to Q and P to Q?

Shalnov [3]

The sum of the distances of R to Q and P to Q is 9 units.

Solution:

Given points are P(–3, 6), Q(3, 6) and R(3, 3).

Distance between two points formula:

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance from R to Q:

Here $x_1=3, y_1=3, x_2=3, y_2=6$

Substitute these in the given formula, we get

Distance = $\sqrt{(3-3)^2+(6-3)^2}$

$=\sqrt{0+9}$

= 3 units

Distance from R to Q is 3 units.

Distance from P to Q:

Here $x_1=-3, y_1=6, x_2=3, y_2=6$

Substitute these in the given formula, we get

Distance = $\sqrt{(3-(-3))^2+(6-6)^2}$

$=\sqrt{36+0}$

= 6 units

Distance from P to Q is 6 units.

Sum of the distances = 3 + 6 = 9 units

Hence the sum of the distances of R to Q and P to Q is 9 units.

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3 years ago

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inysia [295]

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2 years ago