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Alex Ar [27]
3 years ago
10

What is the solution set for 5x−5=15, given the replacement set {0, 1, 2, 3, 4}?

Mathematics
2 answers:
klio [65]3 years ago
8 0
5(4) – 5 = 15

So the answer is A) x = 4
leva [86]3 years ago
8 0

Answer:

Your answer should be A.  {x = 4} hope this helped

You might be interested in
The area of a square garden is 50m^2. How long is the diagonal?
abruzzese [7]

Answer:

Adjacent length = √(50) m

Pythagorean theorem:

a^2 + b^2 = c^2

a = √(50) m

b = √(50) m

c^2 = √(50)^2 + √(50)^2

c^2 = 50 + 50

c^2 = 100

c = 10 m

Diagonal = 10 m

3 0
3 years ago
Consider the oriented path which is a straight line segment L running from (0,0) to (16, 16 (a) Calculate the line integral of t
aalyn [17]

This question is missing some parts. Here is the complete question.

Consider the oriented path which is a straight line segment L running from (0,0) to (16,16).

(a) Calculate the line inetrgal of the vector field F = (3x-y)i + xj along line L using the parameterization B(t) = (2t,2t), 0 ≤ t ≤ 8.

Enter an exact answer.

\int\limits_L {F} .\, dr =

(b) Consider the line integral of the vector field F = (3x-y)i + xj along L using the parameterization C(t) = (\frac{t^{2}-256}{48} ,\frac{t^{2}-256}{48} ), 16 ≤ t ≤ 32.

The line integral calculated in (a) is ____________ the line integral of the parameterization given in (b).

Answer: (a) \int\limits_L {F} .\, dr = 384

              (b) the same as

Step-by-step explanation: <u>Line</u> <u>Integral</u> is the integral of a function along a curve. It has many applications in Engineering and Physics.

It is calculated as the following:

\int\limits_C {F}. \, dr = \int\limits^a_b {F(r(t)) . r'(t)} \, dt

in which (.) is the dot product and r(t) is the given line.

In this question:

(a) F = (3x-y)i + xj

r(t) = B(t) = (2t,2t)

interval [0,8] are the limits of the integral

To calculate the line integral, first substitute the values of x and y for 2t and 2t, respectively or

F(B(t)) = 3(2t)-2ti + 2tj

F(B(t)) = 4ti + 2tj

Second, first derivative of B(t):

B'(t) = (2,2)

Then, dot product between F(B(t)) and B'(t):

F(B(t))·B'(t) = 4t(2) + 8t(2)

F(B(t))·B'(t) = 12t

Now, line integral will be:

\int\limits_C {F}. \, dr = \int\limits^8_0 {12t} \, dt

\int\limits_L {F}. \, dr = 6t^{2}

\int\limits_L {F.} \, dr = 6(8)^{2} - 0

\int\limits_L {F}. \, dr = 384

<u>Line integral for the conditions in (a) is 384</u>

<u />

(b) same function but parameterization is C(t) = (\frac{t^{2}-256}{48}, \frac{t^{2}-256}{48} ):

F(C(t)) = \frac{t^{2}-256}{16}-\frac{t^{2}-256}{48}i+ \frac{t^{2}-256}{48}j

F(C(t)) = \frac{2t^{2}-512}{48}i+ \frac{t^{2}-256}{48} j

C'(t) = (\frac{t}{24}, \frac{t}{24} )

\int\limits_L {F}. \, dr = \int\limits {(\frac{t}{24})(\frac{2t^{2}-512}{48})+ (\frac{t}{24} )(\frac{t^{2}-256}{48})  } \, dt

\int\limits_L {F} .\, dr = \int\limits^a_b {\frac{t^{3}}{384}- \frac{768t}{1152} } \, dt

\int\limits_L {F}. \, dr = \frac{t^{4}}{1536} - \frac{768t^{2}}{2304}

Limits are 16 and 32, so line integral will be:

\int\limits_L {F} \, dr = 384

<u>With the same function but different parameterization, line integral is the same.</u>

6 0
3 years ago
What is 6-4? really hard pls help​
Margarita [4]
The answer is 2! Hope that helps!
8 0
2 years ago
Trapezoid JKLM is shown on the coordinate plane below:
natulia [17]
<span>trapezoid JKLM is translated according to the rule (x, y) → (x + 8, y − 3)
</span><span> the coordinates of point L' are
</span> (-2, -5) → (-2 + 8, -5 − 3)
 (6, -8)
 the answer is 
<span> d. (6, −8)</span>
8 0
2 years ago
Read 2 more answers
Computer output from a regression analysis is provided. Coefficients: Estimate Std. Error t value p-value (Intercept) 7.2960 14.
Crazy boy [7]

Answer:

Step-by-step explanation:

Hello!

You need to test the hypothesis that the slope of the regression is cero.

I've run in the statistic software the given data for Y and X and estimated the regression line:

Yi= 7.82 -1.60Xi

Where

a= 7.82

b=-1.60

Sb= 3.38

The hypothesis is:

H₀: β = 0

H₁: β ≠ 0

α: 0.05

This is a two-tailed test, the null hypothesis states that the slope of the regression is cero, this means that if the null hypothesis is true, there is no linear regression between Y and X.

The statistic for this test is a Student-t

t= <u>  b - β  </u> ~t_{n-2}

      Sb

The critical values are:

Left: t_{n-2; \alpha /2} = t_{2; 0.025} = -4.303

Right: t_{n-2; \alpha /2} = t_{2; 0.975} = 4.303

t= <u>-1.60 - 0 </u>= -0.47

      3.38

the p-value is also two-tailed, you can calculate it by hand:

P(t ≤ -0.47) + (1 - P(t ≤ 0.47) = 0.3423 + (1 - 0.6603) =0.6820

With the level of significance of 5%, the decision is to not reject the null hypothesis. This means that the slope of the regression is equal to cero, i.e. there is no linear regression between the two variables.

I hope this helps!

6 0
3 years ago
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