Find the value of 35° 9

n August 31 of the current year, the assets and liabilities of Gladstone, Inc. are as follows: Cash $27,900; Supplies, $900; Equ

frez [133]

The amount of stockholders’ equity as of August 31 of the current year is $26400.

<h3>How to calculate the equity?</h3>

The owner's equity will be:

= Cash + Supplies + Equipment - Account payable

= 27900 + 900 + 8500 - 7300

= 26400

Therefore, the amount of stockholders’ equity as of August 31 of the current year is

Learn more about equity on:

brainly.com/question/1957305

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8 0

2 years ago

Jada list these fractions that are all equivalent to 1/2:2/4,3/6,4/8,5/10 she notices that each time the numerator increases by

r-ruslan [8.4K]

yes the pattern Jada sees will continue.

Step-by-step explanation:

Im not sure how to explain it but I hope this helps!

8 0

3 years ago

What is -15x=-240?????????

Vlad [161]

You divide -15 on both sides to keep the x alone
-240/ -15=16
A negative divided by a negative is a positive so x=16

4 0

3 years ago

Read 2 more answers

The length of the base of a triangle is twice its height. If the area of the triangle is 16 square kilometers, find the height.

ValentinkaMS [17]

Answer:

4

Step-by-step explanation:

Area of a triangle = bh/2

B = 2h

16 = 2h × 2/2

h×2 = 16

2h = 16

* put a square root on both sides then u will get the answer which is

Height of the triangle = 4

3 0

2 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

forsale [732]

Answer:

a)

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

b)

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

Step-by-step explanation:

The formula is:

$v=\frac{2 \pi R}{T}$

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

$a_c=\frac{4 \pi^{2} R}{T^2}$

Now,

a)

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

$v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97$

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

$a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}$

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

b)

At 30.0° north of the equator:

$R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6$

Now,

Speed = $v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79$

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

$a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}$

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

4 0

3 years ago

Find the value of 35° 9 ​

Find the value of 35° 9