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3 years ago

In 25 words or fewer, compare the mathematical and algebraic expressions.

Mathematics

1 answer:

DedPeter [7]3 years ago

5 0

Answer:

use own words

Step-by-step explanation:

An expression is just a mathematical phrase. In this tutorial, you'll learn about two popular types of expressions: numerical and algebraic expressions. A numerical expression contains numbers and operations. An algebraic expression is almost exactly the same except it also contains variables.

in mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number)

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What is the slope and y-intercept of the equation: y=1/5x-1

Alekssandra [29.7K]

Answer:

The slope is 1/5= rise/run

and y-intercepts is -1

Step-by-step

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3 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 $0.127 \frac{cm}{min}$

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4 years ago

under a dilation, the point (2,6) is moved to (6,18) what is the scale factor of the dilation? enter your answer in the box

Anestetic [448]

Step-by-step explanation:

The equation of line passing through the two points is $\frac{y-y_{1} }{x-x_{1} }=\frac{y_{2} -y_{1}}{x_{2} -x_{1}}$

When substituted,the equation becomes $\frac{y-6}{x-2}=\frac{18-6}{6-2}$

which when simplified is $y=3\times x$

The line clearly passes through origin.

The distance between two points is $\sqrt{(y_{1}-y_{2} )^{2}+(x_{1}- x_{2} )^{2}}$

Distance between origin and $(2,6)$ is $\sqrt{40}$ .

Distance between origin and $(6,18)$ is $\sqrt{360}$

Scale factor is $\frac{distance\text{ }of\text{ }p_{2}\text{ from origin}}{distance \text{ } of \text{ } p_{1}\text{ from origin}}$

So,scale factor is $\frac{\sqrt{360} }{\sqrt{40} }$

which when simplified becomes 3.

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3 years ago

In 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B. If 70 houses wer

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Answer:

The number of houses built in Town B is 56.

Step-by-step explanation:

We are given that in 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B.

Also, 70 houses were built in Town A during 210.

Let the number of houses built in Town B be 'x'.

So, according to the question;

Number of houses built in Town A = Number of houses built in Town B + 25% of the houses built in Town B

$70 = x + (25\% \times x)$

$70 = x(1+0.25)$

$x=\frac{70}{1.25}$

x = 56

Hence, the number of houses built in Town B is 56.

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3 years ago

Find the perimeter. Use 3.14 for pi.

balu736 [363]

Step-by-step explanation:

perimeter=4×6-1/4 ×2π×3=24-1/2×3.14×3=24-4.71=19.29cm

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3 years ago