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3 years ago

6

Please help. I am having trouble with this question. This is sixth-grade mathematics and is worth 50 points. I will mark brainli

est, rate 5 stars, and heart the answer if it is correct.

2 answers:

antiseptic1488 [7]3 years ago

8 0

Answer:

x is greater than or equal to 3

d

Verdich [7]3 years ago

5 0

Answer:

LMOoooiwgtf*dwtfiykd&sew$aertryuiop{:kokhujgyudtreswagsrdcfgbhjkml:<

Step-by-step explanation:

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Find the value of X and y from the figure

Shalnov [3]

Answer:

180° = x + (2x -15)

180°=3x-15

3x = 195

x =65°

x + y = 180°

65+ y= 180°

y = 115°

8 0

3 years ago

Read 2 more answers

What is the slope of the line that contains the points (7,-1) and (-2,2)?

Vesnalui [34]

Answer: -1/3

(7,-1)= (x1,y1)

(-2,2)= (x2,y2)

Slope= $\frac{y2-y1}{x2-x1}$

= $\frac{2-(-1)}{-2-7}$

= $\frac{1}{-3}$

1/-3 is the same thing as -1/3

8 0

3 years ago

F(4)=13 , f(0)=21<br><br> write an equation for the linear function f with the giver values

Fudgin [204]

Is there a worksheet

3 0

3 years ago

Y= -4/3x - 5/2<br><br>y-intercept:<br><br>slope:

CaHeK987 [17]

Answer:the answer is Y= -4/3x - 5/2

Step-by-step explanation:

3 0

3 years ago

Suppose a change of coordinates T:R2→R2 from the uv-plane to the xy-plane is given by x=e−2ucos(5v), y=e−2usin(5v). Find the abs

anzhelika [568]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

Step-by-step explanation:

From the question we are told that

$x = e^{-2a} cos (5v)$

and $y = e^{-2a} sin(5v)$

Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as

$| \frac{\delta (x,y)}{\delta (u, v)} | = | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |$

$= |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right] |$

$Let \ a = -2e^{-2u} cos(5v), \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)$

So

$\frac{\delta (x,y)}{\delta (u, v)} | = |det \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | a * b - c* d |$

substituting for a, b, c,d

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 e^{-4u} (cos^2 (5v) + sin^2 (5v))|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

7 0

3 years ago