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Rashid [163]
3 years ago
5

How many solutions does the nonlinear system of equations graphed below have?

Mathematics
2 answers:
Nata [24]3 years ago
8 0

Answer:

The answer is the option A

Four

Step-by-step explanation:

we know that

The solutions of the nonlinear system of equations shown in the graph are equal to the intersection points both graphs

In this problem

The intersection points both graphs are four points

therefore

The systems has four solutions

see the attached figure to better understand the problem

Artemon [7]3 years ago
4 0
I think its 4 because the lines intersect 4 times but its been too long since ive done nonlinear equations
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andrew11 [14]

Answer:

1st one

Step-by-step explanation:

4 0
3 years ago
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A repair bill for a car is $648.45. The parts cost $265.95. The labor cost is $85 per hour. Write and solve an equation to find
zysi [14]

Answer:

( t - p ) /  85x =  h

( t - 265.95 ) /  85x  = h

648.45 - 265.95 / 85x = 3.32h

h = 3.32

Therefore ( t - p ) /  85x =  3.32

Step-by-step explanation:

648.45 - 265.95

282.50 /85

= 3.32 hrs

Finding equation to find hours can be shown as

( t - p ) /  85x =  3.32

How we found equation to find total

Parts = 19 x 14 - 5/100 = 266- 0.05

648.45 - 265.95 = 282.5

282.5/85 = 3.323  near to 3,233529

Labour = 85x  = 17 (5 + x)

Equation = 17(5+ x) + 14(19) - 5/100 = 648.45 where x = 85

Equation = ( 85x x 3  1/3) + 265 - 19/20 = t

3 0
2 years ago
Area of composite shapes!!
Bingel [31]

Answer:

34.5 units²

Step-by-step explanation:

7 x 3 = 21

4.5 x 3 = 13.5

21 = 13.5 = 34.5 units²

4 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

7 0
3 years ago
The 154 tenth-graders at Wilson High School were polled on whether they enjoyed their algebra or geometry course more. The resul
wolverine [178]

Answer: 80/154

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Step-by-step explanation:

8 0
3 years ago
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