PLZ NEED HELP i will GIVE YOU BRAINS FOR THE BEST EXPLINATION :)

Mathematics

2 answers:

Llana [10]2 years ago

5 0

The answer is –7/26 (in decimal form –0.2692307)

r-ruslan [8.4K]2 years ago

4 0

the Answer:

here :) -7/26

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Help!! I need help!

Tasya [4]

Answer:

Step-by-step explanation:

-5x-49>113

+49 +49

———————-

-5x>162

—- ——

-5 -5

x<32.4

6 0

2 years ago

Taryn conducted a science experiment on saturation. She added sugar to sugar water solution at different intervals. The graph sh

dusya [7]

Equation of a line:
$y=mx+c$

m = gradient: The difference between two y points and two x points.
$m= \frac{\Delta y}{\Delta x}$

c = y-intercept: Where the line crosses the y-axis (x=0)

You have:
$y=\textunderscore\textunderscore \ x + \textunderscore$
so you are missing the m and the c.

To calculate m find two y coordinates -you have (12, 7) and (0, 1)- and subtract them. Then divide this by the subtracted values of the x coordinates -you have (12, 7) and (0, 1)- This gives:

$m= \frac{7-1}{12-0}$
$m= \frac{6}{12}$
$m=0.5$

To calculate the c, you just see where the line crosses the y-axis. Because you have the point (0, 1), you know that when x=0, y=1. Because x=0 is on the y-axis, you can tell that the line passes through y=1. This makes your c = 1:

$c=1$

When you plug these values into the equation you get your answer:

$y=0.5x+1$

6 0

3 years ago

Use cylindrical coordinates. find the volume of the solid that lies within both the cylinder x2 y2 = 9 and the sphere x2 y2 z2 =

marissa [1.9K]

In Cartesian coordinates, the region is given by $-3\le x\le3$ , $-\sqrt{9-x^2}\le y\le\sqrt{9-x^2}$ , and $-\sqrt{16-x^2-y^2}\le z\le\sqrt{16-x^2-y^2}$ . Converting to cylindrical coordinates, using

$\begin{cases}\mathbf x(r,\theta,\zeta)=r\cos\theta\\\mathbf y(r,\theta,\zeta)=r\sin\theta\\\mathbf z(r,\theta,\zeta)=\zeta\end{cases}$

we get a Jacobian determinant of $r$ , and the region is given in cylindrical coordinates by $0\le\theta\le2\pi$ , $0\le r\le3$ , and $-\sqrt{16-r^2}\le z\le\sqrt{16-r^2}$ .

The volume is then

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{z=-\sqrt{16-r^2}}^{z=\sqrt{16-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\dfrac{4(64-7\sqrt7)\pi}3$