What is the best estimate for 82% of 503?

George is 4 times Dorothy's age. If George is 56, how old is Dorothy? Let d = Dorothy's age.

Yuri [45]

4d=56
d=14. Ad a result, Dorothy is 14 years old. Hope it help!

3 0

2 years ago

Read 2 more answers

Please help and solve for x! (: thank you in advance.

Vera_Pavlovna [14]

First look at the given numbers, especially 123 degrees which if you add up the other side it equals 180 degrees so:
180-123= 57
Ok now you know two sides of the triangle and you know that all the 3 sides add up to 360 degrees so:
57+75= 132
180-132= 48

6 0

3 years ago

I WILL GIVE YOU BRAINLIEST!!

ollegr [7]

Answer:

32 goldfish

Step-by-step explanation:

Given:

Number of fish tanks = 8
Volume of water per fish tank = 5,544 in³
Water required by 1 goldfish = min 1,386 in³

Number of fish per tank = water per tank ÷ water per goldfish

= 5544 ÷ 1386

= 4

Therefore, each tank can hold a maximum of 4 goldfish.

Greatest number of goldfish = number of tanks × fish per tank

= 8 × 4

= 32

3 0

1 year ago

PLEASE HELP ASAP PLEASE

emmainna [20.7K]

Answer:

The equation that represents the line passing through the point (2, -4) with a slope of one half is

$f(x) = \frac{1}{2}x - 5$

Step-by-step explanation:

The equation of a line can be described by a first order equation in the following format:

$f(x) = ax + b$

In which a is the slope of the line.

Solution:

The line slope is $\frac{1}{2}$ , so $a = \frac{1}{2}$ .

The equation of the line now is:

$f(x) = \frac{1}{2}x + b$

The problem states that the line passes through the point(2,-4). This means that when x = 2, f(x) = -4

So:

$f(x) = \frac{1}{2}x + b$

$-4 = \frac{1}{2}*(2) + b$

$-4 = 1 + b$

$b = -5$

So, the equation that represents the line passing through the point (2, -4) with a slope of one half is

$f(x) = \frac{1}{2}x - 5$

3 0

2 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

2 years ago