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2 years ago

Which of the following points are in the first quadrant of the xy-plabe? check all that apply

Mathematics

2 answers:

Serga [27]2 years ago

8 0

Answer: B and D because for the 1st quadrant both x and y need to be positive

Step-by-step explanation:

Readme [11.4K]2 years ago

7 0

B. (2,2)

D. (4,8)

The format for 1st quadrant is: (+,+).

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Factor to the given polynomial completely. 4x³+24x²-288x

STatiana [176]

Answer: $\Large\boxed{4x(x+12)(x-6)}$

Step-by-step explanation:

Given expression

4x³ + 24x² - 288x

Factorize 4x out from the expression

4x · x² + 4x · 6x - 4x · 72

4x (x² + 6x - 72)

Cross multiply to factorize the remaining polynomial expression

The meaning is to allow the factored product of the constant to add up to the first-degree term

x 12

x -6

Combine the result

$\Large\boxed{4x(x+12)(x-6)}$

Hope this helps!! :)

Please let me know if you have any questions

3 0

1 year ago

Read 2 more answers

Calculate the flux of the vector field F⃗ (x,y,z)=(exy+9z+4)i⃗ +(exy+4z+9)j⃗ +(9z+exy)k⃗ through the square of side length 3 wit

ikadub [295]

The square (call it $S$ ) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is $\vec n=\vec\imath-\vec k$ , which is enough information to figure out the equation of the plane containing $S$ :

$(x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x$

We can parameterize this surface by

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

for $0\le x\le\frac3{\sqrt2}$ and $0\le y\le3$ . Then the flux of $\vec F$ , assumed to be

$\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k$ ,

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\vec F(\vec s(x,y))\cdot\vec n\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^3\int_0^{3/\sqrt2}4\,\mathrm dx\,\mathrm dy=\boxed{18\sqrt2}$

3 0

3 years ago

Who could help me with this please:)))))))

8_murik_8 [283]

Q57) if 1/4 of a packet cover 1/5 then using a packet is using 4* 1/4 of it. This means that 4*1/5 of the garden would be covered which = 4/5.

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3 0

3 years ago

What’s 8 divided by 9

Licemer1 [7]

Answer:

0.88888888888

Step-by-step explanation:

3 0

2 years ago

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Which is the correct answer choice?

pav-90 [236]