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anzhelika [568]
3 years ago
13

Please help me! Im really confused on how to solve problems like this

Mathematics
2 answers:
slega [8]3 years ago
7 0
A = -bc
÷-c ÷-c
b = a ÷ -c
nadya68 [22]3 years ago
7 0
To solve this problem, we must isolate b from the rest of the problem.

a= -bc

Firstly, divide -bc by c

a/c=-bc/c

The c on -bc cancels out

a/c=-b

Transfer the negative to he other side of the equation

b= -a/c

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Can someone please help me with the second question….will give brainliest :(
den301095 [7]

Answer:

its 3.

Step-by-step explanation:

4 0
3 years ago
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The area of the surface of the swimming pool is 210 square feet. What is the length d of the deep end (in feet)?
luda_lava [24]
The answer is 3.16 ft

The surface area (SA) of the rectangular prism is:
SA = 2lw + 2ld + 2wd

However, the pool is without the top side, so the area of the top side should be excluded from the surface area of the pool. The surface area of the rectangular pool is:
SA = lw + 2ld + 2wd

We know:
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8 0
2 years ago
Help on this problem?? please?
Reptile [31]

MrBillDoesMath!


Answer to #4:  81/256 * s^8 * t^ 12


Comments:

(7x^3) ^ (1/2)   =  7 ^ (1/2)  *  x^(3/2)   where  ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.

---------------------

(1)             (27s^7t^11)^ (4/3)  

               = 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)


As 27 = 3^3, 27 ^(4/3) = 3^4 = 81

               

(2)  (-64st^2)^ (4/3)  =     (-64)^(4/3) * (s^4/3) * t(^8/3)


As 64 = (-4)^3,  (-64)^(4/3) = (-4)^4 = +256                                        


So (1)/(2) =

81 * s^(28/3)* t^(44/3)

-------------------------------     =

256 s^(4/3) * t^((8/3)


81/256 *  s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =


81/256 * s^(24/3) * t (36/3) =

81/256 * s^8        * t^ 12



MrB

5 0
3 years ago
A rectangular mat has a length of 12 in. and a width of 4 in. Drawn on the mat are three circles. Each circle has a radius of 2
pickupchik [31]

|\Omega|=4\cdot12=48\\ |A|=3\cdot3.14\cdot2^2=37.68\\\\ P(A)=\dfrac{37.68}{48}\approx0.79

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3 years ago
Simplify the following cos(-x)tanx
RSB [31]
\bf \textit{symmetry identites}\\\\
cos(-\theta)=cos(\theta)\\\\
-----------------------------\\\\
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