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2 years ago

The number 1 is a factor of _________.

Mathematics

2 answers:

Marianna [84]2 years ago

6 0

Answer:

The number 1 is a factor of C)all whole numbers.

Masja [62]2 years ago

4 0

<h3>hello!</h3>

1 is a factor of all whole numbers.

Every single number, including odd numbers and even numbers, is divisible by 1 (evenly, without any remainders)

Remember, whole numbers include odd numbers, even numbers, and 0.

Hope everything is clear; if you need any more explanation and/or clarification, kindly let me know, and I will comment and/or edit my answer.

You might be interested in

if i put 1500 into my savings account and earned 180.00 of interest at 4% simple interest , how long was my money in the bank

tresset_1 [31]

T = ?

i = 180.00

R = 4% ; 4/100 => 0.04

p = 1500

i = p x R x t

180.00 = 1500 x 0.04 x t

180.00 = 60 x t

t = 180.00 / 60

t = 3 years

hope this helps!

6 0

3 years ago

A sector with a radius of 15 cm has a central angle measure of 8pie/5

Citrus2011 [14]

Answer:

The area of the sector is $180\pi \ cm^2$ .

Step-by-step explanation:

We have,

Radius of circle is 15 cm

Central angle is $\dfrac{8\pi}{5}\ \text{radian}$

It is required to find the area of the sector. The formula used to find the area of sector is given by :

$A=\dfrac{1}{2}r^2\theta$

Plugging all the values in above formula as :

$A=\dfrac{1}{2}\times 15^2\times \dfrac{8\pi}{5}\\A=180\pi \ cm^2$

So, the area of the sector is $180\pi \ cm^2$ .

3 0

3 years ago

Given two dependent random samples with the following results:

Naddik [55]

Answer:

The null hypothesis will not be rejected.

Step-by-step explanation:

The hypothesis for the test is:

<em>H</em>₀: There is no difference between the two population means, i.e. <em>d</em> = 0.

<em>Hₐ</em>: There is a significant difference between the two population means, i.e. <em>d</em> ≠ 0.

Consider the Excel output attached.

The mean of the differences is, $\bar d=-2.75$ .

The standard deviation of the differences is, $S_{d}=4.268$ .

Compute the test statistic as follows:

$t=\frac{\bar d}{S_{d}/\sqrt{n}}=\frac{-2.75}{4.268/\sqrt{8}}=-1.82$

The degrees of freedom is, n - 1 = 7.

Compute the <em>p</em>-value as follows:

$p-value=2\cdot P(t_{7}$

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level.

<em>p</em>-value = 0.112 > α = 0.02.

The null hypothesis will not be rejected.

4 0

4 years ago

Given: ∠ A = ∠ C, E is the midpoint of AC<br> Prove: △AEB △CED

bija089 [108]

Answer:

Step-by-step explanation:

3.AE = CE E is the midpoint of AC

4. ∠AEB = ∠CED Vertically opposite angles are equal

5. ΔAEB = ΔCED ASA congruence

4 0

3 years ago

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0

3 years ago