Multiply and simplify.

The length of a rectangle is 3 inches more than three times the width. the perimeter is 94 inches. find the length and width.

Zina [86]

Let length be (l) and width be (w).

According to the question, l=3w+3......equation1
And perimeter that is 2(l+w) =94........equation2

Solve these two equations, you get your answer.
l=36
w=11

3 0

2 years ago

Line segments XY and ZY are tangent to circle O. Circle O is shown. Triangle X Y Z has points X and Z on the circle. Lines X Y a

pickupchik [31]

Answer:

The correct option is;

An isosceles triangle

Step-by-step explanation:

The parameters given are;

Line segments XY and ZY are tangents to circle O.

ΔXYZ has points X and Z on the circle.

XY and ZY are tangents that intersect at point Y outside the circle'

Therefore, given the circle has center O, then;

OY ≅ OZ = Radius of circle with center O

∠OXY ≅ ∠OZY = 90° (The radius line from the circle center to the tangent point is perpendicular)

OY ≅ OY (Reflexive property)

ΔOYX and ΔOYZ are right triangles (Triangle with one angle = 90°

ΔOYX ≅ ΔOYZ (Hypotenuse Leg, HL, rule of congruency)

Therefore;

XY = ZY (Corresponding Parts of Congruent Triangles are Congruent CPCTC)

Triangle XYZ is an isosceles triangle (Triangle with two equal sides).

4 0

2 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$

we know that $cos^2A+sin^2A=1$ , so we have that

$f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)$

$lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19$

So f(0,0)=ln19.

8 0

2 years ago

Akinte, Chuck, Jenny, and Norma are all going to college next year.

Licemer1 [7]

Answer and explanation:

Akinye Reynoso

Chuck peningan

Norma van hee

Jenny olowski

Akinte is not going to the University of Memphis nor university of Sierra tech nor university of Penn valley but goes to St Mary just like Reynoso(his surname)

Chuck is not going to St Mary nor Sierra tech nor Penn valley but goes to Memphis just like pennigan

Norman does not go to Memphis nor Penn valley nor St Mary but goes to Sierra tech just like van hee

Jenny does not go to Sierra tech nor St Mary nor Memphis but goes for Penn valley just like olowski

3 0

2 years ago

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