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zloy xaker [14]
2 years ago
14

Can someone plz help me.?

Mathematics
1 answer:
Step2247 [10]2 years ago
5 0
The Answer Is = 164.6
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Help please I don’t understand this at all & been stuck here for a hour
ELEN [110]

Answer:

Step-by-step explanation:

4b + 8 = 20

Subtract 8 from both sides

4b + 8 - 8 = 20 - 8

4b = 12

Now divide both sides by 4

4b/4 = 12/4

b = 3

Check:

LHS = 4b + 8

       = 4*3 + 8

        = 12 + 8

        = 20 = RHS

4 0
3 years ago
What is the area of a rectangle if 1 base is 14 in, the 2nd base is 3.5 in and the height is 8 in?​
algol13

Check the picture below.

notice, we have three pair of sides, let's add them all up to get the area

\stackrel{\textit{left and right}}{2(8\cdot 14)}+\stackrel{\textit{front and back}}{2(3.5\cdot 8)}+\stackrel{\textit{top and bottom}}{2(3.5\cdot 14)}\implies 378~in^2

5 0
2 years ago
Rearrange the equation so q is the independent variable.<br><br> 9q-4=3r-6<br><br> r = __
erik [133]

Answer:r=3q+2/3

Step-by-step explanation:

Step 1: Flip the equation.

3r-6=9q-4

Step 2: Add 6 to both sides.

3r-6+6=9q-4+6

3r=9q+2

Step 3: Divide both sides by 3.

3r/3=9q+2/3

r=3q+2/3

Hopefully it’s right

6 0
3 years ago
The sum of a number x and 5 equals 14.
Fudgin [204]
Sum means addition so the equation is x+5=14

To get the answer, just do 14-5 and the answer is 9
4 0
3 years ago
What is the maximum number of turns for the graph of f(x)=x^4+3x^2?
Stels [109]

Answer:

The turns of a graph is represented by the number of maximum or minimum that the function has.

If we differenciate f(x) we get:

f'(x)=4x^3+6x

f'(x)=2x(2x^2 + 3)

Therefore f'(x) =0, when x=0. Given that negative roots are not defined.

Therefore, the number of turns will be given by the number of solutions of f'(x) which is 1.

Attached you find the graph of the function which confirms the number of turns.

If the function had other solutions, the maximum number of turns it could have is 3! because f'(x) is a third degree polynomial, therefore it can't have more than 3 solutions!

6 0
3 years ago
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