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2 years ago

Which of the following mood words has a negative connotation?

Mathematics

2 answers:

jok3333 [9.3K]2 years ago

8 0

The answer is Mocking. Letter B.

Gre4nikov [31]2 years ago

5 0

Mocking
usually mocking is used like “people are mocking me” , which is basically insulting

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Can some answer these math questions asap thanks

Alexus [3.1K]

Answer:

3.) 63=3x+7y

4.) Y-intercept is 9

Step-by-step explanation:

(For 3) For every time a person gets a touchdown, x represents that

for every time a person person scored a successful extra-point attempt, y represents that.

And they team scores 63, so....

the equation would look like this

63=3x+7y

(For 4) You find the y-intercept by turning the standard form equatio into the slop intercept form

$(-3x)63=3x+7y (-3x)\\(\frac{1}{7}) 7y=(-3x+63)(\frac{1}{7} )\\y=-\frac{3x}{7} +9$

(Helper's note, multiplying by its reciprocal is the same as dividing it by a whole number, in the case dividing the number by 7 is the save as multiplying the number by $\frac{1}{7}$ )

Then plugging x=0 and solve for y, which in this case, y=9

$y=\frac{-3(0)}{7} +9\\y=9$

(For the graphing) Turn 63=3x+7y into a slope intercept form, which is $y=-\frac{3x}{7} +9$ and graph from there

4 0

2 years ago

Convert the integral below to polar coordinates and evaluate the integral.

Lelechka [254]

Answer:

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

Step-by-step explanation:

We are trying to evaluate this integral.

$\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}} xy \,\,dxdy$

The first thing that we have to do is understand this region in the plane.

$\{ (x,y) \in \mathbb{R} : 0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2} \}$

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

$\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4} \,\,\, 0\leq r \leq 4 \}$

Now remember that when we do the polar transformation we use the following formula

$\int\limits_{a}^{b} \, \int\limits_{c}^{d} f(x,y) \,dxdy = \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2} r* f(rcos(\theta),rsin(\theta)) \,drd\theta$

Then our integral would be

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$