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3 years ago

A one-pound five-ounce can of hairspray costs $9.95.

Mathematics

1 answer:

Svet_ta [14]3 years ago

3 0

Answer:

D. $0.47

Step-by-step explanation:

1 pound = 16 ounce

16 + 5 = 21

9.95/21 = 0.47

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Describe the correlation of the scatterplot.

pickupchik [31]

Negative correlation

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3 years ago

Read 2 more answers

What is the range of the equation

a_sh-v [17]

The range of the equation is $y>2$

Explanation:

The given equation is $y=2(4)^{x+3}+2$

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

$y=2 \cdot 4^{x+3}+2$

This can be written as $y=2^{1+2(x+3)}+2$

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

$x=2^{1+2(y+3)}+2$

Solving for y, we get;

$x-2=2^{1+2(y+3)}$

Applying the log rule, if f(x) = g(x) then $\ln (f(x))=\ln (g(x))$ , then, we get;

$\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)$

Simplifying, we get;

$(1+2(y+3)) \ln (2)=\ln (x-2)$

Dividing both sides by $\ln (2)$ , we have;

$2 y+7=\frac{\ln (x-2)}{\ln (2)}$

Subtracting 7 from both sides of the equation, we have;

$2 y=\frac{\ln (x-2)}{\ln (2)}-7$

Dividing both sides by 2, we get;

$y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}$

Let us find the positive values for logs.

Thus, we have,;

$x-2>0$

$x>2$

The function domain is $x>2$

By combining the intervals, the range becomes $y>2$

Hence, the range of the equation is $y>2$

7 0

3 years ago

Absolute value of |x|= 5\6

12345 [234]

Hey there!☺

$Answer:\boxed{x=\frac{5}{6},-\frac{5}{6}}$

$Explanation:$

$|x|=\frac{5}{6}$

To solve for x, simplify both sides of the equation and then isolate the variable:

$|x|=\frac{5}{6}\\x=\frac{5}{6},-\frac{5}{6}$

Hope this helps!☺

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3 years ago

Samuel needs 233 feet of wood to build a fence. The wood comes in lengths of 11 feet. How many total piece of wood will Samuel n

mezya [45]

2563 beause 233×11=2563 pieces of wood

4 0

3 years ago

Jason and Britton are driving to St. George. Jason got a 20 mile head start and drove an

harina [27]

Answer:

2 hours, 150 miles

Step-by-step explanation:

The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.

<h3>Separation distance</h3>

Jason got a head start of 20 miles, so that is the initial separation between the two drivers.

<h3>Closure speed</h3>

Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.

<h3>Closure time</h3>

The relevant relation is ...

time = distance/speed

Then the time it takes to reduce the separation distance to zero is ...

closure time = separation distance / closure speed = 20 mi / (10 mi/h)

closure time = 2 h

Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.

<em>Additional comment</em>

The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.

3 0

1 year ago