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Delicious77 [7]
3 years ago
5

Please answer asap

Mathematics
2 answers:
Artyom0805 [142]3 years ago
7 0
Using the distributive property...you can multiply and add your way to an answer.

6.8(6.7 - 7.2) - 2(4.6 + 1.2)                      (use distributive property)
6.8(6.7) + 6.8(-7.2) - 2(4.6) - 2(1.2)            (multiply)
45.56 - 48.96 - 9.2 - 2.4                        (add)
- 15                                                         (answer)
Juli2301 [7.4K]3 years ago
6 0
6.8(6.7 – 7.2) – 2(4.6 + 1.2) = -15
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How do you solve 26/57= 849/5x
sergeinik [125]
If you would like to solve 26/57 = 849/5x, you can do this using the following steps:

<span>26/57 = 849/5x          /*(5x)
</span>26/57 * (5x) = 849     /*57
26 * 5x = 849 * 57    /26
5x = 48393/26        /5
x = <span>48393/(26*5)
x = 372.25

The correct result would be </span><span>372.25.</span><span>
</span>
8 0
3 years ago
thè speed of a runner increased steadily during the first three seconds of a race. Her speed at half-second intervals is given i
bagirrra123 [75]
undefined

8 0
1 year ago
Read 2 more answers
If the simple interest on $6,000 for 9 years is $3,240 , then what is the interest rate?
sergey [27]

Answer:

6%

Step-by-step explanation:

principal =6000$

time=9years

simple interest=3,240$

now,

Rate=SI ×100/P×T

=3240×100/6000×9

=324000/54000

=6%

6 0
3 years ago
When the smaller of two consecutive integers is added to two times the​ larger, the result is 35. Find the integers.
Aleonysh [2.5K]

x+2(x+1)=35

x+2x+2=35

3x=35-2=33

x=33/3=11

smaller number=11

larger number=11+1=12

4 0
3 years ago
Which simplifications of the powers of i are correct?
OleMash [197]

The correct simplifications of powers of i are:

a. i^27=−i

d. i^8=1

e. i^5=i

Step-by-step explanation:

We know that

i^2=-1\\i^3=-i\\i^4=1

We can use the smaller powers of i to solve the larger powers.

a. i^27=−i

i^{27}\\={i^{24}}.i^3\\={i^{4*6}}.-i\\=(i^4)^6.-i\ \ \ \ As\ we\ know\ i^4=1\\=(1)^6.-i\\=1*-i\\=-i

b. i^28=i

i^{28}\\=i^{4*7}\\=(i^4)^7\\=(1)^7\\=1

c. i^14=1

i^{14}\\=i^{12}.i^2\\=(i^{4*3}).i^2\\=> i^2=-1\\=(i^4)^3.-1\\=(1)^3*-1\\=-1

d. i^8=1

i^8\\=i^{4*2}\\=(i^4)^2\\=(1)^2\\=1

e. i^5=i

i^5\\=i^{2*2*1}\\=i^2*i^2*i\\=(-1)(-1)(i)\\=i

Hence,

The correct simplifications of powers of i are:

a. i^27=−i

d. i^8=1

e. i^5=i

Keywords: Imaginary units, Iota

Learn more about imaginary units at:

  • brainly.com/question/8806598
  • brainly.com/question/8805387

#LearnwithBrainly

6 0
3 years ago
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