Find the slope-intercept equation of the line,

Every morning, siblings Jack and Jill roll a regular six-sided die to determine who will get to ride in the front seat on the wa

Arada [10]

Answer:

3/6

LCM of 6 and 20= 60

3/6=1/2=10/20=10days

so estimated 10 days

5 0

3 years ago

An angle measures 7.4° less than the measure of its complementary angle. What is the measure of each angle?

Xelga [282]

Answer:

41.3° and 48.7°

Step-by-step explanation:

Complementary angles are angles with the sum of 90°

if one of the angle is 7.4° less than other

(90 - 7.4) ÷ 2 = 41.3° is the measure of smaller angle and

41.3 + 7.4 = 48.7° is the measure of bigger angle

5 0

2 years ago

Meghan also wants to walk to get some exercise, rather than going to the gym. She decides to walk along arc AB. How far will she

Karo-lina-s [1.5K]

Answer:

π*AB

Step-by-step explanation:

I m not quite sure about your question because it is lack of details.

However, as my per understanding, Meghan decides to walk along arc AB it means she walks an interval equal to the circumference of the circle AB.

So, apply the formula in caculating the circumference of the circle, we have:

π*d = π*AB

So, she walk a π*AB distance.

8 0

3 years ago

Three children each had the same amount of money in their savings account. One of the children withdrew a quarter of her money a

malfutka [58]

The total money with them was $100 in their accounts

3 0

3 years ago

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) Large and small drains are opened

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) Only the small drain is opened

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$

$t_{s} = 5.303\,h$

The draining time when only the small drain is opened is 5.303 hours.

7 0

3 years ago