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3 years ago

What is a divisibility rules for two

Mathematics

1 answer:

jekas [21]3 years ago

3 0

When the number ends with 2,4,6,8 or 0, the number is divisable by 2

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Simplify: (−2a2b) • (4a5b2)<br> please can someone help!

xz_007 [3.2K]

Hey there hope this helps you out

5 0

3 years ago

HELPPP PLSSS 18. A triangle has verticesA(4,3), B(6,9) and C(2, -3). The triangle is rotated 90°

Mariulka [41]

Answer:

F and H are NOT true

Step-by-step explanation:

Rotation 90° CCW is represented by the transformation ...

(x, y) ⇒ (-y, x)

Then the rotated coordinates are ...

A(4, 3) ⇒ A'(-3, 4) . . . . . quadrant II

B(6, 9) ⇒ B'(-9, 6) . . . . . quadrant II

C(2, -3) ⇒ C'(3, 2) . . . . . quadrant I

The rotated figure will be congruent to the original, because rotation is a rigid transformation.

Consider the answer choices:

F: (-3, -4) is A' --- NOT true

G: triangles are congruent --- true

H: B is in quadrant III --- NOT true

J: (3, 2) is the coordinates of C' --- true

Both choices F and H are NOT true.

8 0

2 years ago

A bag of birdseed that normally cost $22.25 is marked up to $25.30. What is the percent markup ?

LenaWriter [7]

25.3-22.25 = 3.05
What percent of 22.25 is 3.05?
22.25x = 3.05
x = .1370786517
It's marked up 13.7%

4 0

3 years ago

Armando spent $9.70 at the grocery store. he spent $2.95 on a bag of baby spinach and bought tomatoes at $2.25 per pound. How ma

Sever21 [200]

He bought 3 pounds of tomatoes

6 0

3 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago