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3 years ago

(2,8) & (4,60) which one is proportional

Mathematics

1 answer:

tensa zangetsu [6.8K]3 years ago

4 0

They are both proportional. 1:4 and 1:15

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HELP FIRST TO ANSWER ALL IS BRAINLEIST!!!!!!!!!

andreev551 [17]

Answer:

1. D) 1 9/20

2. B) 7/12

4. D) $50.00

Step-by-step explanation:

The working is as follows :

1. Mrs. Lang bought 3/4 pound of red grapes and 7/10 pound of green grapes. To find out how many are in total we have to add both values total.

3/4 + 7/10

The LCM of 4 and 10 is 20.

3/4 * 5 = 15/20

7/10 * 2 = 14/20

15/20 + 14/20 = 29/20 = 1 9/20

2. 1/3 + 1/4

The LCM of 3 and 4 is 12

1/3 * 4 = 4/12

1/4 * 3 = 3/12

4/12 + 3/12 = 7/12

4. Sharon spent 2/3 of her clothing allowance; she was given $75. All we have to do is find 2/3 of $25 :

2/3 * $75 = $50.00

4 0

3 years ago

Expand.<br> Your answer should be a polynomial in standard form.<br> (3a^2 – 1)(-3a² + 5) =

34kurt

Answer:-9a^4+18a^2-5

(3a

−1)(−3a

+5)

=3a

(−3a

+5)−1(−3a

+5)

=−9a

+15a

+3a

−5

=−9a

+18a

−5

4 0

3 years ago

What is the average number of landfills in the US from 2010-2018?

exis [7]

2000???????????!!??!!??!??!??!?!!??!??!??!

7 0

3 years ago

What is 12% of 75? =)))

ankoles [38]

Answer:

Hope this helps!

Step-by-step explanation:

12% of 75 would be 9

5 0

3 years ago

Read 2 more answers

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