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3 years ago

You are cutting as many 32 1/4 inch sections as you can out of the board to make skateboards. how many skateboards can you make?

1 answer:

8 0

all you have to do is get a calculator and divide 32 and 1/4 AND DIVIDE IT into whatever you are planning to cut and whatever you get then multiply that by either 32 and 1/4 ad then you will have your answer sorry if you don't understand or if you get it wrong.

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Find the no term of 6,12,18,24

s2008m [1.1K]

You should try Google it helps

5 0

4 years ago

PLEASE HELP ASAP! I don’t recall how to do this!

MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

$(1.05)^x=1.885$

In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:

$ln(1.05)^x=ln(1.885)$

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

$x=\frac{ln(1.885)}{ln(1.05)}$

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

$ln(x*x^2)=5$

Simplifying gives you

$ln(x^3)=5$

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

$ln(x)=\frac{5}{3}$

Take the inverse ln by raising each side to e:

$e^{ln(x)}=e^{\frac{5}{3}}$

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

$\frac{1}{2}=e^{.1x}$

"Undo" that e by taking the ln of both sides:

$ln(.5)=ln(e^{.1x})$

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

$(2^2)^x-6(2)^x=-8$

Now we will bring over the -8 by adding:

$(2^2)^x-6(2)^x+8=0$

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let $u=2^x$

When we do that, we can rewrite the polynomial as

$u^2-6u+8=0$

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:

$2^x=4,2^x=2$

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

$2^2=2^x$

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

$2^x=2^1$

Now that the bases are the same, we can say that

x = 1

5 0

3 years ago

64 is the sum of 18 and Jenny's score.

AlexFokin [52]

Answer:

64=18+j

64-18=18+j-18

46=j

Jenny's score is 46

7 0

3 years ago

Ed bought 3 liters of water, 2,750 milliliters of sports drinks, and 2.25 liters of juice. For numbers 4a-4e, select True or Fal

Naily [24]

<span>4a) 1000 milliliters = 1 liter, thus 3 liters = 3 x 1000 = 3000 milliliters.
Thus Ed bought 3000 - 2750 = 250 more water than sports drink.
Therefore, Ed bought 250 milliliters more water than sports drink is True.

4b) Ed bought 3 - 2.25 = 0.75 more water than juice.
Therefore, Ed bought 1.25 liters more water than juice is False.

4c) Ed bought 2.25 liters = 2.25 x 1000 = 2250 milliliters of juice.
Thus, Ed bought 2750 - 2250 = 500 more sports drink than juice.
Therefore, Ed bought 50 milliliters more sports drink than juice is False.

4d) Ed bought 2750 milliliters = 2750 / 1000 = 2.75 liters of sports drink.
Ed bought 2.75 - 2.25 = 0.5 liters more sports drink than juice.
Ed bought 0.5 liter more of sports drink than juice is True.

4e) Ed bought 3000 milliliters of water and 2250 milliliters of juice.
Ed bought 3000 - 2250 = 750 more milliliters of water than juice.
Therefore, Ed bought 75 milliliters more water than juice is False.</span>

8 0

4 years ago

There are___puppies that weigh less than 102 g

Nataly [62]

Need more info I can help just do not know what the question is

5 0

4 years ago