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4 years ago

How do I solve this I do not know the steps for this question

Mathematics

1 answer:

Debora [2.8K]4 years ago

5 0

This question is easy the answer is x=-5

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How do you interpret slope and y- intercept from a table?

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Answer:

graph the table and fine the rise over run

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4 years ago

50 POINTS Simplify. 57‾‾√ 35√7 5√7 7√5 12√7

liraira [26]

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5 square root 7

Step-by-step explanation:

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4 years ago

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What are the steps to divide 336/7

melamori03 [73]

Just plug in a calculator 336/7 and you should get a non decimal answer.

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3 years ago

A horse race has 1313 entries and one person owns 44 of those horses. Assuming that there are no ties, what is the probability

777dan777 [17]

Answer: 0.0014

Step-by-step explanation:

Given : Number of entries in horse race = 13

Number of horses owned = 4

We assume that there are no ties .

Then , the probability that those four horses finish first, second ,third ,and fourth is given by :-

$\dfrac{4!}{^{13}P_4}=\dfrac{4!}{\dfrac{13!}{(13-4)!}}\\\\=\dfrac{4!\times9!}{13!}=.0013986013986\approx0.0014$

Hence, the required probability = 0.0014

6 0

4 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago