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GarryVolchara [31]
3 years ago
7

What is the solution to the equation below round your answer to two decimal places 7^x=77

Mathematics
1 answer:
Crank3 years ago
3 0

Answer:

D= 2.23

Step-by-step explanation:

a p e x

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Solve the equation <br>8(4- x) = 7x + 2<br>x=​
Jet001 [13]

Answer: x=2

Step-by-step explanation:

Expand.

32−8x=7x+2

Add 8x to both sides.

32=7x+2+8x

Simplify  7x+2+8x7x+2+8x  to  15x+215x+2.

32=15x+2

Subtract 22 from both sides.

32−2=15x

Simplify  32−2  to  30.

30=15x

Divide both sides by 15.

30/15=x

Simplify 30/15 to 2

2=x

Switch sides.

x=2

7 0
3 years ago
Need help ASAP please and thank you
noname [10]

Answer:

-3m^{12}{n^{6} }

Step-by-step explanation:

To begin simplifying this, we can first divide the coefficients, giving us:

\frac{-3m^{5}n^{4}  }{m^{-7} n^{-2} }

We know that when dividing exponents, this means we need to subtract the exponent on the denominator from the numerator. This gives us:

\frac{-3m^{5+7} n^{4+2} }{1}

Now, simplifying this gets:

-3m^{12}{n^{6} }

8 0
3 years ago
Decide whether the two expressions are equivalent for each problem below.
Vesnalui [34]

Answer:

Step-by-step explanation:

6 0
2 years ago
I need help fasttt with work
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Answer:

-19.9 ft.

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Would appreciate some help with this:<br><br> Verify (sinq+cosq)^2 - 1 = sin2q
Phantasy [73]

Hello from MrBillDoesMath!

Answer:

See Discussion below

Discussion:

(sinq + cosq)^2 =                                     => (a +b)^2 = a^2 + 2ab + b^2

(sinq)^2 + (cosq)^2 + 2 sinq* cosq        => as (sinx)^2 + (cosx)^2 = 1

1 + 2 sinq*cosq              (*)


Setting a = b = q in the trig identity:

sin(a+b) = sina*cosb + cosa*sinb                      

sin(2q) =                      (**)

sinq*cosq + cosq*sinq      => as both terms are identical

2 sinq*cosq


Combining (*) and (**)

(sinq + cosq)^2 = 1 + 2sinq*cosq     => (**) 2sinq*cosq =  sqin(2q)

                          = 1 +  sin(2q)

Hence

(sinq + cosq)^2  = 1 +  sin(2q)            => subtracting 1 from both sides

(sinq + cosq)^2  - 1 =   sin(2q)  

The last statement is what we are trying to prove.



Thank you,

MrB

7 0
3 years ago
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