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Gennadij [26K]
2 years ago
10

Given: f(x)=1-x^2,g(x)=1-2x, and h(x)= 1/x^2+1 find f(g(a+1)) find g(1/h(z))

Mathematics
1 answer:
AleksAgata [21]2 years ago
7 0

Answer:

that boy look like joe jeffereson wide neck a

Step-by-step explanation:

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Find x. (Use pic for more info)
V125BC [204]

Answer:

x=75

Step-by-step explanation:

Hey There!

If you didn't know the exterior angle of a triangle is equal to the sum of the opposite interior angles of a triangle

So essentially what I'm saying is that

2x-5=70+x

step 1 add 5 to each side

-5+5 cancels out

70+5=75

now we have

2x=70+x

step 2 subtract x from each side

x-x cancels out

2x-x=x

we're left with

x=75

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A car for $10,000 and a $8,000 loan is taken. The loan is for 3 years and the interest
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10922 thats the answer

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Which graph represents the inequality y-2x≤1?
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2 years ago
Hi looking for help me
tankabanditka [31]

Answer:

In my own words:

Definition from search: parallel lines are lines in a plane which do not meet; that is, two straight lines in a plane that do not intersect at any point are said to be parallel.

My definition: Two lines that are equally apart from start to end.

These are / / or | | or = or \ \

PARALLEL LINES

6 0
3 years ago
The length l, width w, and height h of a box change with time. At a certain instant the dimensions are l = 3 m and w = h = 6 m,
Gemiola [76]

Answer:

a) The rate of change associated with the volume of the box is 54 cubic meters per second, b) The rate of change associated with the surface area of the box is 18 square meters per second, c) The rate of change of the length of the diagonal is -1 meters per second.

Step-by-step explanation:

a) Given that box is a parallelepiped, the volume of the parallelepiped, measured in cubic meters, is represented by this formula:

V = w \cdot h \cdot l

Where:

w - Width, measured in meters.

h - Height, measured in meters.

l - Length, measured in meters.

The rate of change in the volume of the box, measured in cubic meters per second, is deducted by deriving the volume function in terms of time:

\dot V = h\cdot l \cdot \dot w + w\cdot l \cdot \dot h + w\cdot h \cdot \dot l

Where \dot w, \dot h and \dot l are the rates of change related to the width, height and length, measured in meters per second.

Given that w = 6\,m, h = 6\,m, l = 3\,m, \dot w =3\,\frac{m}{s}, \dot h = -6\,\frac{m}{s} and \dot l = 3\,\frac{m}{s}, the rate of change in the volume of the box is:

\dot V = (6\,m)\cdot (3\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot (3\,m)\cdot \left(-6\,\frac{m}{s} \right)+(6\,m)\cdot (6\,m)\cdot \left(3\,\frac{m}{s}\right)

\dot V = 54\,\frac{m^{3}}{s}

The rate of change associated with the volume of the box is 54 cubic meters per second.

b) The surface area of the parallelepiped, measured in square meters, is represented by this model:

A_{s} = 2\cdot (w\cdot l + l\cdot h + w\cdot h)

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time:

\dot A_{s} = 2\cdot (l+h)\cdot \dot w + 2\cdot (w+h)\cdot \dot l + 2\cdot (w+l)\cdot \dot h

Given that w = 6\,m, h = 6\,m, l = 3\,m, \dot w =3\,\frac{m}{s}, \dot h = -6\,\frac{m}{s} and \dot l = 3\,\frac{m}{s}, the rate of change in the surface area of the box is:

\dot A_{s} = 2\cdot (6\,m + 3\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m+6\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m + 3\,m)\cdot \left(-6\,\frac{m}{s} \right)

\dot A_{s} = 18\,\frac{m^{2}}{s}

The rate of change associated with the surface area of the box is 18 square meters per second.

c) The length of the diagonal, measured in meters, is represented by the following Pythagorean identity:

r^{2} = w^{2}+h^{2}+l^{2}

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time before simplification:

2\cdot r \cdot \dot r = 2\cdot w \cdot \dot w + 2\cdot h \cdot \dot h + 2\cdot l \cdot \dot l

r\cdot \dot r = w\cdot \dot w + h\cdot \dot h + l\cdot \dot l

\dot r = \frac{w\cdot \dot w + h \cdot \dot h + l \cdot \dot l}{\sqrt{w^{2}+h^{2}+l^{2}}}

Given that w = 6\,m, h = 6\,m, l = 3\,m, \dot w =3\,\frac{m}{s}, \dot h = -6\,\frac{m}{s} and \dot l = 3\,\frac{m}{s}, the rate of change in the length of the diagonal of the box is:

\dot r = \frac{(6\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot \left(-6\,\frac{m}{s} \right)+(3\,m)\cdot \left(3\,\frac{m}{s} \right)}{\sqrt{(6\,m)^{2}+(6\,m)^{2}+(3\,m)^{2}}}

\dot r = -1\,\frac{m}{s}

The rate of change of the length of the diagonal is -1 meters per second.

6 0
3 years ago
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