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3 years ago

Please help me with all of the work for this equation: -8t-23=-15 thanks:))))

1 answer:

6 0

Add 23 to both sides of the equal sign giving you -8t= 8 then divide both sides of the equal sign by -8 which gives you t= -1

Here's the work:

<span>-8t - 23 = -15 23 + (-15) = 8
</span> +23 +23
-8t = 8
t= -1

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The difference of two numbers is 192, the sum is 3782

LenaWriter [7]

(3782 - 192) : 2 = 1795 (first number)

1795 + 192 = 1987 ( second number)

--------------------------

1987 - 1795 = 192

1987 + 1795 = 3782

4 0

3 years ago

If there os 150 capacity and 36 os used what os the remaining percentage

Veseljchak [2.6K]

150 - 36 = 114 os remaining;
114/150 = 0,76 (as a decimal) = 76% (as a percentage);

6 0

3 years ago

PLS HELP Choose the solution to this inequality. 43<−83y y < −12 y > 12 y < 43 y > 4

OlgaM077 [116]

Given:

The inequality is:

$\dfrac{4}{3}$

To find:

The solution of the given inequality.

Solution:

We have,

$\dfrac{4}{3}$

Multiply both sides by 3.

$4$

Divide both sides by -8 and change the sign of inequality.

$\dfrac{4}{-8}>\dfrac{-8y}{-8}$

$-\dfrac{1}{2}>y$

It can be written as:

$y$

Therefore, the correct option is A.

8 0

2 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago

Explain the difference between a unit rate and a percentage

love history [14]

Answer:

Ratio is the proportion between two numbers.

Percent is the ratio times 100.

Rate is the change of a number per unit time.

Step-by-step explanation:

1.A rate refers to the frequency by which a certain event happens while a ratio refers to the relationship between the size, number, or degree of two or more things.

2.A rate is a comparison between two measurements of the same units while a ratio is the proportion of one thing to another.

3.A rate refers to the fixed quantity of two things while a ratio refers to the relationship between various things.

4.A ratio indicates the difference between things while a rate indicates the changes in their measurements or units.

5.A ratio is indicated by the quotient of one quantity divided by the other while a rate is indicated by the comparison between two things.

Hope it helped ! Have a nice day

6 0

2 years ago