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1 year ago

Please answer fast The vertices of a rectangle are located at the coordinates (1, 4), (1, 5), (6, 5),

Mathematics

1 answer:

hammer [34]1 year ago

3 0

Answer:

See below ~

Step-by-step explanation:

Drawing the rectangle (Refer attachment)

Horizontal sides

There are two x-values present : 1 and 6
Find the difference
6 - 1 = 5
The horizontal sides of the rectangle are 5 units long

Vertical sides

Two y-values are present : 4 and 5
Find the difference
5 - 4 = 1
The vertical sides of the rectangle are 1 unit long

Perimeter

2(Horizontal side + Vertical side)
2(5 + 1)
2(6)
12
The perimeter of the rectangle is 12 units

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Combine like terms to create an equivalent expression 3x²+5y²-x^2+2y^2-4

Alja [10]

Answer

3x²+5y²-x^2+2y^2-4

2x^2+7y^2-4

7 0

2 years ago

HELP i need it on my final

ANTONII [103]

Answer:

-2

Step-by-step explanation:

when x = 5, f(x) = -2

hope this helps. good luck

8 0

2 years ago

Please Help! I need help to my math question

dem82 [27]

First you need to get rid of the parenthesis by distributing the 0.6 to each term inside.
(0.6)(10n) + (0.6)(25) =10+5n
6n +15 = 10+5n subtract the 5n on both sides and subtract 15 from both sides
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5 0

3 years ago

Read 2 more answers

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

Given that MN = 5, NO = 12, and MO = 13, find cos O.

Galina-37 [17]

Answer:

12/13

Step-by-step explanation:

Given that MN = 5, NO = 12, and MO = 13, find cos O.

Since the reference angle is P, hence;

MN is the opposite = 5

MO is the hypotenuse = 13 (longest side)

NO is the adjacent = 12

Cos O = adj/hyp

Substitute the given values

Cos O = 12/13

Hence the value of Cos O is 12/13

8 0

2 years ago