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Gekata [30.6K]
3 years ago
14

How many solutions does the following system of linear equations have?

Mathematics
1 answer:
kari74 [83]3 years ago
7 0

Answer:

No Solution

Step-by-step explanation:

because both of them have same gradient or slope.

we can say that those linear equations are parallel

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What equation(s) represent joint variations? check all that apply c=2pi z=3x/y y=3x+2 w=abc/4 v=lwh
Alinara [238K]

Answer:

Step-by-step explanation:

Joint variations occurs when one variable depends on the value of two or more variables. The variable varies directly or indirectly with the other variables combined together. The other variables are held constant. From the given examples, the equation(s) that represent joint variations are

1) z = 3x/y

z varies directly with x and inversely with y.

2) w = abc/4

w varies inversely with a,b and c. 4 is the value of the constant of variation.

7 0
3 years ago
Determine the input value for which the statement
adelina 88 [10]

Answer:

The x-value at which the two functions' values are equal is 5.

Step-by-step explanation:

Let f(x) = 3 and g(x) = x-2, if f(x) = g(x), then:

x - 2 = 3

x = 3+2

x = 5

The x-value at which the two functions' values are equal is 5.

8 0
3 years ago
A. solve differential equa y' = 4x square root (1-y^2)
Allisa [31]

Step-by-step explanation:

a)

dy/dx = 4x √(1 − y²)

Separate the variables:

dy / √(1 − y²) = 4x dx

Integrate:

sin⁻¹ y = 2x² + C

Solve for y:

y = sin(2x² + C)

b)

Plug in initial value.

4 = sin(0 + C)

Sine cannot be greater than 1, so there is no solution.

3 0
3 years ago
An educational organization in California is interested in estimating the mean number of minutes per day that children between t
Hatshy [7]

Answer:

The critical value for a 98% CI is z=2.33.

The 98% confidence interval for the mean is (187.76, 194.84).

Step-by-step explanation:

We have to develop a 98% confidence interval for the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day.

We know the standard deveiation of the population (σ=21.5 min.).

The sample mean is 191.3 minutes, with a sample size n=200.

The z-value for a 98% CI is z=2.33, from the table of the standard normal distribution.

The margin of error is:

E=z\cdot \sigma/\sqrt{n}=2.33*21.5/\sqrt{200}=50.095/14.142=3.54

With this margin of error, we can calculate the lower and upper bounds of the CI:

LL=\bar x-z\cdot\sigma/\sqrt{n}=191.3-3.54=187.76\\\\\\UL=\bar x+z\cdot\sigma/\sqrt{n}=191.3+3.54=194.84

The 98% confidence interval for the mean is (187.76, 194.84).

4 0
3 years ago
Which of the following would be a good argument to lease rather than buy?
Arturiano [62]

Answer: B. I prefer the warranty that covers repair costs of a new car

5 0
4 years ago
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