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3 years ago

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One day a store sold 36 sweatshirts. White ones cost $10.95 and yellow ones cost $11.50. In all, $404.65 worth of sweatshirts

were sold. How many of each color were sold?
The store sold ____ white sweatshirts

1 answer:

Hatshy [7]3 years ago

7 0

Answer:

17 white sweatshirts were sold and 19 yellow sweatshirts were sold.

Step-by-step explanation:

Let <em>w</em> represent the number of white sweatshirts sold and <em>y</em> represent the number of yellow sweatshirts sold.

We can write a system of equations to represent the situation.

Since the store sold a total of 36 sweatshirts, the sum of the white and yellow sweatshirts must total 36. So:

$y+w=36$

And since each white sweatshirt cost $10.95 and each yellow sweatshirt cost $11.50 and the total profit was $404.65:

$10.95w+11.5y=404.65$

Solve the system. I'll use substitution this time (though elimination will work just as perfect). From the first equation, subtract <em>w</em> from both sides:

$y=36-w$

Substitute this into the second:

$10.95w+11.5(36-w)=404.65$

Distribute:

$10.95w+414-11.5w=404.65$

Simplify:

$-0.55w=-9.35$

Divide both sides by -0.55:

$w=17$

So, 17 white sweatshirts were sold.

Using the modified equation, substitute:

$y=36-(17)=19$

Therefore, 17 white sweatshirts were sold and 19 yellow sweatshirts were sold.

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Solve the equation; 3a/4-2a/3=7/4

DiKsa [7]

Answer: " a = 21 " .

_____________________

<u>Step-by-step explanation</u>:

Given: 3a/4-2a/3 = 7/4 ; Solve for "a" ;

Rewrite as: (3a/4) - (2a/3) = (7/4) ;

Now, for each of the three (3) "denominator values" in "fraction form" within the equation given:

→ Find the "LCD" ["Least Common Denominator"]:
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that is: "3" and "4" ;
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To find the LCD: First; multiply the denominators: "4 * 3 = 12" .
So; the value "12" could be the LCD; so, the value for the LCD is no greater than "12" ; however, there <u><em>could </em></u>be a smaller value.
To determine the LCD:
List the multiples of the given denominators:
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3: 3, 6, 9, <u><em>12</em></u>, 15 .... ;
4: 4, 8, <u><em>12</em></u>, 16... ;
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We find that "12" is, in fact, the LCD of "3" and 4:
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We can multiply each side of the equation by "12" ; to eliminate the "fractional values" :
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$12*[\frac{3a}{4} - \frac{2a}{3}] = 12*[{\frac{7}{4}]$
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<u>Note the</u><u> "</u><u>distributive property</u><u>"</u><u> of multiplication</u>:
→ a(b+c) = ab + ac ;

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As such:
$12*[\frac{3a}{4} - \frac{2a}{3}] = 12*[{\frac{7}{4}]$ ;

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Let us start with the "left-hand side" of the equation:
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$12*[\frac{3a}{4} - \frac{2a}{3}]$ ;

= $[12*\frac{3a}{4}] + [-12 * \frac{2a}{3}]$ ;
= $[12*\frac{3a}{4}] - [12 * \frac{2a}{3}] ;$
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Note: " $[12*\frac{3a}{4}]$ " ;

$= \frac{12}{1} * \frac{3a}{4}$ ;

→ The "12" cancels to a "3" ; and the "4" cancels to a "1" ;
since: "12÷4 = 3" ; and since: "4÷4 = 1" ;
→ and we can rewrite the "left-hand-side" expression as:
→ " $\frac{3}{1} * \frac{3a}{1}$ " ; which we can simplify as:

→ "3 * 3a" ; which we can simplify as: " 9a " .
then we have: " [12 * $\frac{2a}{3}$ ] " ;

which equals:
= " $\frac{12}{1} * \frac{2a}{3}$ " ;
<u>Note</u>: The "12" cancels out to a "4"; & the "3" cancels out to a "1" ;

→ {since: "(12 ÷ 3 = 4)"; & since: "(3 ÷ 3 = 1)" ;
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→ And we can rewrite the expression as:
→ " $\frac{4}{1} *\frac{2a}{1}$ " ; which we can simplify as:
→ " 4 * 2a " ; which we can simplify/calculation as: " 8a " ;
Now, we can rewrite the expression of the "left-hand side"
of the equation as:
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→ " [9a] − [8a] " ; (don't forget to carry down the "minus sign"!) ;
which we can simplify/calculate to get:
→ " [9a − 8a] " ; which we can further simplify/calculate;

→ to get:
→ " 1a " ; or: "a" —the value for which we wish to solve!
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Now, let us examine the "right-hand side" of the equation:
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→ " $\frac{12}{1} * \frac{7}{4}$ " ;
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And we can rewrite the expression as:
→ " $\frac{3}{1} * \frac{7}{1}$ " ; which we can simplify as:
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→ which is the correct answer:
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2 years ago

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OleMash [197]

Answer:

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3 years ago

A pipe is leaking at the rate of 16 fluid ounces per minute. Use unit analysis to find out how many gallons the pipe is leaking

larisa [96]

Answer:

The answer is 7.5 gallons per hour.

Step-by-step explanation:

16 fluid ounces multiplied by 60 minutes equals 960 fluid ounces

16 * 60 = 960

960 fluid ounces divided by 128 fluid ounces (which is 1 gallon) equals 7.5 gallons.

960 / 128 + 7.5

So the answer is 7.5 gallons per hour.

Hope this helps! :)

3 0

2 years ago

Consider the two functions:

koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is $x=\frac{40}{4+5a}$

b) The value of the functions at the point where they intersect is $\frac{10 (28 + 15 a)}{4 + 5 a}$

c) The partial derivative of f with respect to $x$ is $\frac{\partial f}{\partial x} = -5a$ and the partial derivative of f with respect to $a$ is $\frac{\partial f}{\partial x} = -5x$

d) The value of $\frac{\partial f}{\partial x}(3,2) = -10$ and $\frac{\partial f}{\partial a}(3,2) = -15$

e) $\upsilon_1=-\frac{3}{4} = -0.75$ and $\upsilon_2=-\frac{3}{4} = -0.75$

f) equation $\upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14}$ and $\upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}$

Step-by-step explanation:

a) In order to find the $x$ we just need to equal the equations and solve for $x$ :

$f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}$

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of $a$ ) must be the same.

$f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}$

and for $g(x)$ :

$g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}$

c) $\frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a$

$\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x$

d) Then evaluating:

$\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10$

$\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15$

e) Substituting the corresponding values:

$\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40} = -\frac{3}{4} = -0.75$

$\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40} = -\frac{3}{4} = -0.75$

f) Writing the equations:

$\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }$

$\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }$

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4 years ago

Help please!!!!!!!!!!!!!

svp [43]

Answer:

30 is the perimeter i think 2 x 5 = 10 + 2 x 4.5 = 10 x 2 = 20 so 20 + 10 = 30

I think the area is 45, 5 x 4.5 = 22.5 and 2 x 22.5 = 45

Wait for more responses if needed please.

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3 years ago