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kogti [31]
2 years ago
10

The length of a rectangular rug is 4 less than twice its width. The perimeter of the rug is 40 feet. What is the area of the rug

?
1) 96 sq ft
2) 56 sq ft
3) 104 sq ft
4) 86 sq ft
Mathematics
1 answer:
disa [49]2 years ago
6 0
96 sq ft is the answer— here’s my work btw

You might be interested in
X + 2y – 4z = 1<br> -2x – 4y + 4z = 6<br> -4x – 3y – 4z = 21
elena55 [62]

Answer:

x = - 1

y = - 3

z = -2

Step-by-step explanation:

Please see steps in the image attached here.

6 0
3 years ago
Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp
shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If \phi: G \longrightarrow G is an homomorphism, then it must hold

that b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1},

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)

which tells us that \phi is a homorphism. The kernel of \phi

is the set of elements g in G such that g^{n}=1. However,

\phi is not necessarily 1-1 or onto, if G=\mathbb{Z}_6 and

n=3, we have

kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}

(c) If z_1,z_2 \in \mathbb{C}^{\times} remeber that

|z_1 \cdot z_2|=|z_1|\cdot|z_2|, which tells us that \phi is a

homomorphism. In this case

kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}, if we write a

complex number as z=x+iy, then |z|=x^2+y^2, which tells

us that kern(\phi) is the unit circle. Moreover, since

kern(\phi) \neq \{1\} the mapping is not 1-1, also if we take a negative

real number, it is not in the image of \phi, which tells us that

\phi is not surjective.

(d) Remember that e^{ix}=\cos(x)+i\sin(x), using this, it holds that

\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)

which tells us that \phi is a homomorphism. By computing we see

that  kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \} and

Im(\phi) is the unit circle, hence \phi is neither injective nor

surjective.

7 0
3 years ago
Please help if you can!
Mariulka [41]
^^^^^^^^^^^^^^^^^^^^^↑^^^^^^^ try downloading photo math , its helps you alot!!

8 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
What is the LCM of 13/16a^3b^3, 11/30a^5b
vichka [17]

Answer:

The L.C.M of given numbers is \frac{143}{240}a^5b^3.

Step-by-step explanation:

The given expressions are

\frac{13}{16}a^3b^3

\frac{11}{30}a^5b

The factors of given expressions are

\frac{13}{16}a^3b^3=\frac{13}{8}\times \frac{1}{2}\times a\times a\times a\times b\times b\times b

\frac{11}{30}a^5b=\frac{11}{15}\times \frac{1}{2}\times a\times a\times a\times a\times a\times b

L.C.M.=\frac{13}{8}\times \frac{1}{2}\times a\times a\times a\times b\times b\times b\times \frac{11}{15}\times a\times a

L.C.M.=\frac{143}{240}\times a^5\times b^3

L.C.M.=\frac{143}{240}a^5b^3

Therefore, the L.C.M of given numbers is \frac{143}{240}a^5b^3.


8 0
3 years ago
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