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3 years ago

Standard form to slope intercept form 6x+5y=50

2 answers:

6 0

To put it in slope intercept form, you want to solve for y.
5y = -6x + 50
y = -6/5x + 10

So, your slope is -6/5 (or -1.2) and your y-intercept is +10.

Oksana_A [137]3 years ago

5 0

Hope this helps you y+10

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8x – 6 = 2 (4x + 3) <br><br> Is it no solution?

loris [4]

Answer:

X=12

Step-by-step explanation:

8×-6=2(4×+3)

8x-6=8×+6

8×-8×=6+6

×=12

4 0

3 years ago

Read 2 more answers

Identify the equation of the circle that has its center...

Nat2105 [25]

Answer:

Option C

Step-by-step explanation:

Standard equation:

r^2 = (x-h)^2 + (y-k)^2

Radius: r

Center: (h,k)

Plug the values in.

r^2 = (x+8)^2 + (y-15)^2

Using distance formula, you conclude that the radius is 17 since the distance from (-8,15) to (0,0) is 17.

r^2 = 289

That means option C is the right answer.

289 = (x+8)^2 + (y-15)^2

Please mark for Brainliest!! :D Thanks!!

For any questions or more information, please comment below and I'll respond as soon as possible.

8 0

3 years ago

80. A certain insect has a mass of 75 milligrams.

pickupchik [31]

Since 1 milligram = 0.001 grams you just have to multiply 75 milligram by 0.001 grams in order to get the grams.
75 x 0.001= 0.075 grams so the answer is F.

6 0

3 years ago

What is the formula of volume ofba rhombic prism

Dimas [21]

$Volume=the\ base\ area\ multiplied\ by\ the\ height\\\\if\ the\ base\ is\ a\ rhombus,\ \ then\ the\ base\ area= \frac{1}{2}\cdot e\cdot f \ \ \\\\and\ \ \ e,f\rightarrow\ the\ length\ of\ the diagonals\ of\ a\ rhombus\\\\V=\frac{1}{2}\cdot e\cdot f \cdot h\\\\and\ \ h\rightarrow\ the\ height\ of\ prism$

8 0

4 years ago

Read 2 more answers

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under

Shkiper50 [21]

The work is equal to the line integral of $\vec F$ over each line segment.

Parameterize the paths

from (0, 0, 0) to (2, 0, 0) by $\vec r_1(t)=t\,\vec\imath$ with $0\le t\le2$ ,
from (2, 0, 0) to (2, 4, 1) by $\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k$ with $0\le t\le1$ ,
from (2, 4, 1) to (0, 4, 1) by $\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k$ with $0\le t\le2$ , and
from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$

Then the total work done by $\vec F$ over the particle's path is 46.

8 0

3 years ago