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Nataly_w [17]
3 years ago
15

What is the best way to solve this

Mathematics
1 answer:
Komok [63]3 years ago
6 0
Make 2 equations one for the price and one for quantity.  so first set x = AA and y= AAA. for price 37= 1x+.75y. now for the quantity we have x+y=42. Now we can solve by subtracting the equation for price from the equation for quantity and we get .25y =5. solving this we get y= 20. Now we plug that value back in and solve for x. 22 double A's and 20 triple A's.
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⅓(2x-4)+5=-⅔(x+1)<br><br><br>HALP
Kay [80]

\frac{2}{3} x -  \frac{4}{3}  + 5 =  \frac{ - 2}{3} x -  \frac{2}{3 }  \\  \frac{2}{3} x -  \frac{11}{3}  =  -  \frac{2}{3} x -  \frac{2}{3}  \\ 2x + 11 =  - 2x - 2 \\ 4x =  - 13 \\ x =  \frac{ - 13}{4}
6 0
3 years ago
Since it is given that AB ≅ AC, it must also be true that AB = AC. Assume ∠B and ∠C are not congruent. Then the measure of one a
neonofarm [45]
Answer should be AB≠AC.
6 0
3 years ago
Convert 8π9/9
Zolol [24]

Answer:

tell me if i need to resend it

Step-by-step explanation:

5 0
3 years ago
Let f(x) = cos(x) + 2x.<br> Where does f have critical points?
adell [148]

Answer:

<em>f</em> has no critical points.

Step-by-step explanation:

We are given:

f(x)=\cos(x)+2x

A function has critical points whenever its derivative equals 0 or is undefined.

Differentiate the function:

f'(x)=-\sin(x)+2

Since this will never be undefined, solve for its zeros:

0=-\sin(x)+2

Hence:

\displaystyle \sin(x)=2

Recall that the value of sine is always between -1 and 1.

Thus, no real solutions exist.

Therefore, <em>f</em> has no critical points.

6 0
3 years ago
What two numbers can add to 4 but multiple to -12
olasank [31]

Answer:

<em>Numbers: 6 and -2</em>

Step-by-step explanation:

<u>Equations</u>

This question can be solved by inspection. It's just a matter of factoring 12 into two factors that sum 4. Both numbers must be of different signs and they are 6 and -2. Their sum is indeed 6-2=4 and their product is 6*(-2)=-12.

However, we'll solve it by the use of equations. Let's call x and y to the numbers. They must comply:

x+y=4\qquad\qquad [1]

x.y=-12\qquad\qquad [2]

Solving [1] for y:

y=4-x

Substituting in [2]

x(4-x)=-12

Operating:

4x-x^2=-12

Rearranging:

x^2-4x-12=0

Solving with the quadratic formula:

\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

With a=1, b=-4, c=-12:

\displaystyle x=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(-12)}}{2(1)}

\displaystyle x=\frac{4\pm \sqrt{16+48}}{2}

\displaystyle x=\frac{4\pm 8}{2}

The solutions are:

\displaystyle x=\frac{4+ 8}{2}=6

\displaystyle x=\frac{4- 8}{2}=-2

This confirms the preliminary results.

Numbers: 6 and -2

7 0
3 years ago
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