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1 year ago

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A rectangular pool is 20 feet wide and 50 feet long. A deck used for sunning surrounds the pool. The deck is the same width all

the way around the pool. The total area of the deck is 456 square feet. How wide is the walkway?
A. 2 feet
B. 3 feet
C. 4 feet
D. 5 feet

1 answer:

fredd [130]1 year ago

3 0

The answer is B. 3 feet

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You get a large ice-cream cone for $2.99 and a bottle of water for $1.07 . You pay with a 10 -dollar bill. How much change will

Igoryamba

10 - 2.99 = 7.01 and 7.01 - 1.07 = 5.94 so you will get 5.94 in change

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1 year ago

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Mixing two kinds of candy the price of which was $2 and $4 per pound, Ella got a 10-lb mix of candy, which cost $2.90 per pound.

topjm [15]

1. You have the following information given in the problem above:

- Ella mixed<span> two kinds of candy the price of which was $2 and $4 per pound.
- Ella got a 10-lb mix of candy, which cost $2.90 per pound.
</span>
2. Therefore, let's call:

x: pounds of the first kind of candy.
y: pounds of the second kind of candy.

3. Then, you have:

2x+4(10-x)=(2.90)(10)

4. When you clear x, you obtain:

2x+40-4x=29
-2x=29-40
x=-11/-2
x=5.5 pounds

x+y=10
y=10-x
y=10-5.5
y=4.5 pounds

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2 years ago

The box plot shows the number of years during which 24 colleges have participated in a local college fair: A boxplot is titled Y

Anna007 [38]

Answer:

its is actually 4

Step-by-step explanation:

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2 years ago

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What is 2 (log Subscript 3 Baseline 8 + log Subscript 3 Baseline z) minus log Subscript 3 Baseline (3 Superscript 4 Baseline min

Ilya [14]

Answer:

Value of expression in single logarithm is $\log_3\left(2z^2\right)$ .

Step-by-step explanation:

Given expression is,

$2\left(\log_3\left(8\right)+\log_3\left(z\right)\right)-\log_3\left(3^4-7^2\right)$

Now using logarithmic rule to solve the expression as follows,

Applying product rule of logarithmic,

$\log_c\left(a\right)+\log_c\left(b\right)=\log_c\left(ab\right)$

Therefore,

$2\log_3\left(8z\right)-\log_3\left(3^4-7^2\right)$

Applying power rule of logarithmic,

$a\log_c\left(b\right)=\log_c\left(b^a\right)$

Therefore,

$\log_3\left(\left(8z\right)^2\right)-\log_3\left(3^4-7^2\right)$

$\log_3\left(\left(64z^2\right)\right)-\log_3\left(3^4-7^2\right)$

Applying quotient rule of logarithmic,

$\log_c\left(a\right)-\log_c\left(b\right)=\log_c\left(\frac{a}{b}\right)$

Therefore,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{3^4-7^2}\right)$

Simplifying,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{81-49}\right)$

$\log_3\left(\dfrac{\left(64z^2\right)^2}{32}\right)$

$\log_3\left(2z^2\right)$

Therefore value of expression is $\log_3\left(2z^2\right)$

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2 years ago

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What is the horizontal asymptote for y(t) for the differential equation dy dt equals the product of 2 times y and the quantity 1

marta [7]

First, we need to solve the differential equation.
$\frac{d}{dt}\left(y\right)=2y\left(1-\frac{y}{8}\right)$
This a separable ODE. We can rewrite it like this:
$-\frac{4}{y^2-8y}{dy}=dt$
Now we integrate both sides.
$\int \:-\frac{4}{y^2-8y}dy=\int \:dt$
We get:
$\frac{1}{2}\ln \left|\frac{y-4}{4}+1\right|-\frac{1}{2}\ln \left|\frac{y-4}{4}-1\right|=t+c_1$
When we solve for y we get our solution:
$y=\frac{8e^{c_1+2t}}{e^{c_1+2t}-1}$
To find out if we have any horizontal asymptotes we must find the limits as x goes to infinity and minus infinity.
It is easy to see that when x goes to minus infinity our function goes to zero.
When x goes to plus infinity we have the following:
$$$\lim_{x\to\infty} f(x)$$=y=\frac{8e^{c_1+\infty}}{e^{c_1+\infty}-1} = 8$
When you are calculating limits like this you always look at the fastest growing function in denominator and numerator and then act like they are constants.
So our asymptote is at y=8.

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1 year ago