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Scorpion4ik [409]
3 years ago
12

The graph below shows the solution to which system of inequalities

Mathematics
2 answers:
Slav-nsk [51]3 years ago
8 0

Answer: A

explanation:

from the given graph we see that shaded portion lies above the dotted line which shows open interval (y >-3 ).  graph lies below the line y= -x ,hence the solution region will be y ≤ -x ,

hence the solution region is given by

y> -3  and y≤ -x

Oksi-84 [34.3K]3 years ago
5 0
The answer is
the option A
<span>The graph shows the solution of the system of inequalities
y> -3
y<=-x

using a graph tool
see the attached figure</span>

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Whats the volume of the cylinder? Round your answer to the nearest hundred the top is B the left side of the cylinder is 13cm th
Wittaler [7]

Answer:

The volume of the cylinder is 163.3 m³.

Step-by-step explanation:

Given that,

Height = 13 cm

Base = 4 cm

So, radius = 2 cm

We need to calculate the volume of the cylinder

Using formula of cylinder

V=\pi r^2h

Where, r = radius

h = height

Put the value into the formula

V=\pi\times(2)^2\times13

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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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3 years ago
Please. I need help answering these questions
Goryan [66]
If you're struggling with graphing problems, I'd highly recommend that you check out Desmos and Mathaway. All you have to do is type the equations and it graphs it for you. Here's the first question: 

4 0
3 years ago
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Answer:

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\frac{BC}{EF} =\frac{AC}{DF}.

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To solve the proportion, we'll cross multiply by multiplying numerator with denominator across the equal sign.

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Use the concept of Lagrange multipliers.
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