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Alisiya [41]
3 years ago
7

3 - 2(b - 2) = 2 - 7b b = Please show work

Mathematics
1 answer:
Firdavs [7]3 years ago
5 0
3- 2b +4 = 2-7b
3+4 = 2 - 5b
7 = 2 -5b
5 = -5b
-5b /-5 = -1
b = -1
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8. Shelley and Jim sold tickets to the basketball game. Good seats were $3 each and not so good seats cost
OverLord2011 [107]

Answer:

100 not so good seats

50 good seats

Step-by-step explanation:

Let x represent good seats and y represent not so good seats

[i] 3x + 2y = 350

[ii] x+y = 150

[iii] y = 150 - x

substitute this to equation [i]

3x + 2y = .350

3x + 2 (150-x ) = 350

3x + 300 - 2x = 350

3x-2x = 350 - 300

x = 50 good seats

now substitute what you got to equation [iii]

y = 150 - 50 = 100 not so good seats

when we check the answer by using equation [i] we find that it is correct:

3x + 2y = 350

3(50) + 2(100) = 350

150 + 200 = 350

350 = 350

3 0
3 years ago
Please Help! I'm stuck on this question. Show all work!
Valentin [98]

Add all the displacement vectors together, then find the magnitude of this vector sum.

\vec r_1=(72\,\mathrm{km})(\cos60^\circ\,\vec\imath+\sin60^\circ\,\vec\jmath)

\vec r_2=(48\,\mathrm{km})(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)

\vec r_3=(100\,\mathrm{km})(\cos150^\circ\,\vec\imath+\sin150^\circ\,vec\jmath)

Summing these vectors gives the resultant displacement

\vec r=\vec r_1+\vec r_2+\vec r_3\approx(-50.6\,\vec\imath+64.4\,\vec\jmath)\,\mathrm{km}

and we have

\|\vec r\|=\sqrt{(-50.6)^2+64.4^2}\,\mathrm{km}\approx\boxed{81.9\,\mathrm{km}}

so the plane ends up about 81.9 km away from its starting position.

6 0
4 years ago
5. Consider the contexts the follow.
TEA [102]
The answer is C because it’s the one that makes most sense
8 0
3 years ago
Find the area of a circle with a diameter of 5 cm.
sergeinik [125]

Answer:

19.635

Step-by-step explanation:

A=pie Radius^2

7 0
3 years ago
Read 2 more answers
The answers to 13 and 20 and how there done.
ratelena [41]

The sum and product of two functions f(x) and g(x), i.e. (f+g)(x) and (f\cdot g)(x) respectively, are simply defined as the sum and product of the expressions that define each function.

So, as for exercise 13, we have

(f\cdot g)(x) = f(x)\cdot g(x) = (2x^3-5x^2)(2x-1) = 4 x^4- 12 x^3 +5 x^2

Similarly, as for exercise 20, we have

(g + f)(t) = g(t) + f(f) = (2t+5) + (-t^2+5) = -t^2 + 2t + 10

3 0
3 years ago
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