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3 years ago

Mathhh helpp me plzzz

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

4 0

Answer:

170 bbdbdbdbebsbsbsndnd

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Help me please with this

AysviL [449]

Since both angle are congruent (equal/same), which means that both side is equal to 40 degrees, we will put the equation like this:

5x - 5 = 40

Now we’ll solve for x:

5x - 5 = 40
+ 5 + 5 (Add 5 to both sides)

5x = 45
—- —— (Five both side by 5)
5 5

x = 9

Now we left with x, so x is equal to 9.

8 0

3 years ago

On a cold winter moming in Colorado, the outdoor temperature was 5 degrees Fahrenheit at sunrise For the next sex hours the temp

Temka [501]

Answer: 6

Step-by-step explanation:

because i said so

8 0

3 years ago

Which expressions are equivalent to when x0? Check all that apply.

Genrish500 [490]

We have that

$\frac{(x+4)}{3} / \frac{6}{x} = \frac{x*(x+4)}{3*6} \\ \\ = \frac{( x^{2} +4x)}{18}$

therefore

case a)
$\frac{(x+4)}{3} * \frac{x}{6}$
Is equivalent

case b)
$\frac{6}{x} * \frac{(x+4)}{3}$
Is not equivalent

case c)
$\frac{x}{6} * \frac{(x+4)}{3}$
Is equivalent

case d)
$\frac{(2 x^{2} +4x)}{6}$
Is not equivalent

case e)
$\frac{(2 x^{2} +4x)}{18}$
Is equivalent

Hence

the answer is

$\frac{(x+4)}{3} * \frac{x}{6}$

$\frac{x}{6} * \frac{(x+4)}{3}$

$\frac{(2 x^{2} +4x)}{18}$

3 0

3 years ago

Read 2 more answers

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

3 years ago

What is the least common multiple of 10 and 6

laila [671]

30 is the least common multiple

4 0

3 years ago

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