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Lostsunrise [7]

3 years ago

6

You set up a lemonade stand your profit is equal to your revenue from lemonade sales minus your cost to operate the stand your c

ost is eight dollars how many cups of lemonade or must you sell to earn a profit of $30

1 answer:

Ksivusya [100]3 years ago

6 0

Yettt it’s 64 hope this helps

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Two numbers multiply together to make 8, but add together to make - 6 What are the two numbers?

Greeley [361]

Answer:

-4 and -2

Step-by-step explanation:

First find all the factors of 8, that would be 1 and 8, -1 and -8, 2 and 4, and -2 and -4.

Then, adding the groups of numbers you have should show you what would equal -6. So 1 + 8 = 9, -1 + (-8) = -9, same vice versa, 2 + 4 = 6, -2 + (-4) = -6. So -2 and -4 would be your two numbers.

8 0

3 years ago

3x - 1/5= x+1/3 Could you please solve this as soon as possible

VARVARA [1.3K]

x = 0.27

:)

good luck

4 0

3 years ago

What is the solution to the inequality -6( p - 8) < -12?

coldgirl [10]

Answer:

p > 10

Step-by-step explanation:

Given

- 6(p - 8) < - 12

Divide both sides by - 6, reversing the symbol as a result of dividing by a negative quantity.

p - 8 > 2 ( add 8 to both sides )

p > 10

3 0

3 years ago

Read 2 more answers

DNA TAC ACC TTG GCG ACG ACT Write the mRNA strand then the Amino Acid sequence.

Ugo [173]

AUG UGG AAC CGC UGC AGU

7 0

3 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

4 years ago