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shutvik [7]
3 years ago
15

Find the missing information for each trapezoid. Be sure that you are writing the formula and showing work for each step.

Mathematics
1 answer:
Rus_ich [418]3 years ago
8 0
If you have any questions on what I did, just leave a comment.

I tried to add my second image but it won't allow me. So, the base is 6 that you're missing for question 6. Good luck.

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12 – 7x ≤-2 <br> solve and show work
vova2212 [387]

Answer:

x≥2

Step-by-step explanation:

Let's solve your inequality step-by-step.

12−7x≤−2

Step 1: Simplify both sides of the inequality.

−7x+12≤−2

Step 2: Subtract 12 from both sides.

−7x+12−12≤−2−12

−7x≤−14

Step 3: Divide both sides by -7.

−7x/−7 ≤ −14/−7

x≥2

8 0
2 years ago
I need help i don’t know the answer
Julli [10]
Answer:

3x - 4 and 2x + 15

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
the local bakery uses 1.75 cups of flour in each batch of cookies. the bakery usee 5.25 cups of flour this morning. How many bat
Travka [436]

Answer:

Using 5.25 cups of flour would make 3 batches of cookies. He made 180 cookies.

Step-by-step explanation:

To see how many batches you make all you do is take the 5.25 and divide it by 1.75 because thats how much it takes to make 1. When you divide it you get 3. If there are 5 dozen in each batch thats 60 cookies because 1 dozen is 12. Then,since he made 3 batches you multiply the 60 by 3 and you get 180.

Hope this helps!

4 0
3 years ago
How can you minimize the risk from investment?
adoni [48]
Don’t invest too much into it
3 0
3 years ago
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1. Make sure your answers and work is SHOWN please and steps are in order.
JulijaS [17]

Been a while since I've done synthetic division.

1a. Let's assume that's supposed to be an equals sign

p(x) = 2x⁴ - 3x³ - 6x² + 5x + 6

possible rational roots have factors of 6 in the numerator and of 2 in the denominator.  We'll only worry about negative numerators.

Factors of six: 1,2,3,6, and we don't forget -1,-2,-3,-6

Factors of 2: 1,2

Possible rational roots:

(dividing by 1:) 1,-1,2,-2,3,-3,6,-6

(dividing by 2:) 1/2, -1/2  (2/2=1 is a duplicate, don't have to repeat it), 3/2, -3/2

Possible rational roots: 1,-1,2,-2,3,-3,6,-6, 1/2, -1/2, 3/2, -3/2

Synthetic division, trying x=1,

   1 | 2  -3  -6  5  6

             2  -1  -7  -2

       2    -1  -7  -2  4

Got a remainder of 4, so 1 isn't a root;

Trying x=-1  

-1 | 2  -3  -6  5  6

         -2   5  1  -6

     2  -5  -1  6  0

Zero remainder, found a root, x=-1.  This division says

(2x⁴ - 3x³ - 6x² + 5x + 6) / (x + 1) = 2x³ - 5x² - x + 6

Same set of rational roots on the cubic, we continue with x=2

2 | 2 -5 -1 6

         4 -2 -6

    2  -1 -3 0

Another zero remainder, x=2 is a root.  We're left with 2x² - x - 3 = 0, which factors as

(2x - 3)(x + 1) = 0

That's a second factor of x+1.  Our final factorization is

2x⁴ - 3x³ - 6x² + 5x + 6 = (x + 1)²(x-2)(2x - 3)

Fourth degree with a positive leading coefficient so goes to +infinity at both ends.  Double zero at x=-1, so it's tangent there, just touching the x axis, then down through x=3/2 and up through x=2.  

I'll leave the actually sketching and the other two polynomials to you -- that took some time.

4 0
3 years ago
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