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The cost of 12 pounds of apples costs $35.50. What is equation for the cost in dollars, d, for the cost of a pound, p of apples

eduard

Answer:

12p = 35.50d

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

8 0

3 years ago

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Number between 49 and 95 that's a multiple of 5,6 and 10

fiasKO [112]

5,10,15,20,25,30,35,40,45,50,55,60
6,12,18,24,30,36,42,48,54,60
10,20,30,40,50,60
The answer is 60

8 0

3 years ago

Find L when P= 24 and W = 6.<br> P = 2L+ 2W (perimeter of a rectangle)<br> L =

Anon25 [30]

Answer:

6

Step-by-step explanation:

24 = 2L + 2(6), multiply...

24 = 2L + 12, subtract 12 to both sides...

12 = 2L, divide 2 to both sides to isolate the varibale L

6 = L

Therefroe, L = 6

8 0

3 years ago

By selling a radio for $8400 a dealer gained 12% .how much money did she gain

Mkey [24]

See, in these type of questions always first assume your original price to be x. And now try to write whatever information is given to you.

Now, you have that it has been sold for 12% profit, so basically it is sold at. (1 + (12/100))x = 1.12x

Now it is given that if it had been sold for 33₹ more. I.e. if it had been sold at. 1.12x + 33

Net gain would have been 14%.

So gain of 14%, means the cost to be (1+(14/100))x = 1.14x

And basically both 1.12x +33 and 1.14x are final costs, so

1.14x = 1.12x + 3

=> 1.14x - 1.12x = 3

=> 0.02x = 3

=> x = 3÷0.02

=> x= 1650₹

So original cost is 1650₹.

Hope you understand the solution and the way to tackle these type of problems if not then let me know.

hope it helps

pls mark my anwers as brainiest

6 0

3 years ago

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Find the multiplicative inverse of 6 + 2i

Marina CMI [18]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2774989

________________

Find the multiplicative inverse of

$\mathsf{z=6+2i}$

________

The inverse multiplicative of $\mathsf{z=a+bi}$ is

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\qquad\quad(a\ne 0~~and~~b\ne 0)}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi}}\\\\\\
=\mathsf{\dfrac{1\cdot (a-bi)}{(a+bi)\cdot (a-bi)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-\,\diagup\hspace{-10}abi+\,\diagup\hspace{-10}abi-(bi)^2}}$

$=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot i^2}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot (-1)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2+b^2}}\\\\\\\\
\therefore~~\mathsf{\dfrac{1}{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}\,i\qquad\quad\checkmark}$

________

For this question,

$\mathsf{z=6+2i}$

So,

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{6+2i}}\\\\\\
=\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\
=\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\
=\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}$

$\therefore~~\mathsf{\dfrac{1}{z}=\dfrac{3}{20}-\dfrac{1}{20}\,i}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

6 0

3 years ago

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