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lianna [129]
3 years ago
6

the length of a picture frame is 3 inches greater than the width. The perimeter is less than 52 inches. Describe the dimensions

of the frame.
Mathematics
1 answer:
bonufazy [111]3 years ago
5 0
Hi. To solve this problem, we break it down like this:

P = 2L + 2W < 52

Set it equally

L + W = 26 (because 1/2 of 52 is 26)
(W + 3) + W = 26
2W = 23
W = 11 1/2
L = W + 3 = 14 1/2

So the final dimensions are: Width = 11 1/2 and the Length is 14 1/2

I hope this helps.

Take care,
Diana
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A 3-gallon bottle of fabric softener costs $33.24. What is the price per quart?
asambeis [7]

Answer:

$2.77 per quart.

Step-by-step explanation:

There are four quarts in a gallon.

4 * 3 = 12

There are 12 quarts in 3 gallons.

\frac{price}{quart} =\frac{33.24}{12}=\frac{33.14/12}{12/12}=\frac{2.77}{1}=\boxed{2.77}

The price per quart should be $2.77.

Hope this helps.  

8 0
3 years ago
Classify the triangle by its side lengths.
Harrizon [31]

Answer:

this triangle is an acute triangle

3 0
3 years ago
PLZ SHOW ALL WORK
lara31 [8.8K]

Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width:  w = 12 - 2x
length: l = 18 - 2x
height: h = x

Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x

Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6

Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160

Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224

And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.
3 0
3 years ago
Which one of the following formulas will find the sum of interior angles of a polygon with n sides
Yakvenalex [24]
<span><span>Each time we add a side (triangle to quadrilateral, quadrilateral to pentagon, etc), we </span>add another 180°<span> to the total of the interior angle.</span></span>  
Triangle, sum interior angle is 180
Square, is 360
Pentagon is 540  
So the formula is <span>180 ( n – 2 )</span>
7 0
3 years ago
Read 2 more answers
Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

7 0
3 years ago
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