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3 years ago

NEED QUICKLY!!!! -Please solve the image below

Mathematics

1 answer:

AnnyKZ [126]3 years ago

4 0

Yes,this is a doozy.

   <u>-3 (y - 5) = 15</u>

Eliminate the parentheses on the left side:

   -3(y) -3(-5) = 15

   -3y + 15 = 15

Subtract 15 from each side:

   -3y = 0

Divide each side by -3 :

      <em> y = 0</em>

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What is the range of the function on the graph?all real numbersO all real numbers less than or equal to -1all real numbers less

Yuki888 [10]

The range of the function f(x) is the set of all values that function f takes.

The domain of the function f(x) is the set of all possible values for x.

From the given graph you can see that the domain is all real numbers, $x\in (-\infty,\infty).$

The maximal y-value that f takes is 3 at x=-1. For all another x from the domain, y is less than 3.

Thus, the range of the given function is $y\in (-\infty,3].$

Answer: $y\in (-\infty,3].$

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2 years ago

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The what test is used to determine whether a function is one to one

olga2289 [7]

Answer:

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Step-by-step explanation:

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3 years ago

A drop of oil makes a circular ring in a water puddle. The radius of the oil ring expands at 1.5 cm per second. If r(t)=1.5t and

Readme [11.4K]

Answer:

$a(t) =n(0.25t^2)$

Step-by-step explanation:

Given

$r(t) = 0.5t$

$a(r) = nr^2$

Required

Determine the area in terms of time

From the question, we have:

$r(t) = 0.5t$ -> radius per time

and

$a(r) = nr^2$ --> area from radius

The interpretation of the question is to find the composite function: a(t)

If $a(r) = nr^2$ , and $r(t) = 0.5t$ , the

Substitute 0.5t for r

$a(t) =n(0.5t)^2$

$a(t) =n*0.25t^2$

$a(t) =n(0.25t^2)$

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2 years ago

Log4 216-log4 8 =log4 x

tekilochka [14]

216/8= 27. X= 27. Subtract you divide

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3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

2 years ago