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3 years ago

What is 10x+3-6x simplify

Mathematics

2 answers:

uranmaximum [27]3 years ago

6 0

The answer is 4x+3.

Hope this helps ;)

Mademuasel [1]3 years ago

4 0

4x + 3

Hope this helped :)

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2(5x + 3) -x + 3 = 18

Dvinal [7]

Answer:

x = 1

Step-by-step explanation:

7 0

3 years ago

What two-digit numbers greater than 69 are divisible by 2 and 11?

Mariulka [41]

88... 88/11=8 and 88/2=44

8 0

3 years ago

2.) Fill in the blank.

Montano1993 [528]

Answer:

E.) 1

Step-by-step explanation:

Firstly we will solve for L.H.S.

L.H.S. = $Csc^2\theta$

Since we know that $Csc^2\theta$ is the inverse of $Sin^2\theta$ .

So we can say that;

$csc^2\theta=\frac{1}{sin^2\theta}$

Now For R.H.S.

$Cot^2\theta+1$

Since we can rewrite $cot^2\theta$ as $\frac{cos^2\theta}{sin^2\theta}$ .

Now we can say that the R.H.S. is;

$\frac{cos^2\theta}{sin^2\theta}+1$

Now we add the fraction and get;

$\frac{cos^2\theta+sin^2\theta}{sin^2\theta}$

Now according to trigonometric identity;

$cos^2\theta+sin^2\theta=1$

So, $\frac{cos^2\theta+sin^2\theta}{sin^2\theta}=\frac{1}{sin^2\theta}$

Here,

$csc^2\theta=\frac{1}{sin^2\theta}$ and $Cot^2\theta+1$ = $\frac{1}{sin^2\theta}$ <em> </em>

L.H.S. = R.H.S.

Hence $csc^2\theta=cot^2\theta+1$

4 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

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

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

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

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

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

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

The function y=x²-4x+8 approximates the height, y, of a bird, and its horizontal distance, x, as it flies from one

trapecia [35]

When the bird is 2 feet away from the first fence post, it reaches its minimum height of 4 feet , Minimum at (2,4)

Therefore Option C and D is the correct answer.

<h3>What is a Quadratic Function?</h3>

The equation for a quadratic function is given by

y = a(x-h)² +k²

where (h,k) is the extreme value(vertex).

Here the equation given is

y = x²-4x +8

y = x²-4x+4+4

y = (x-2)² +4

Here the value of a = 1 , h = 2 and k = 4

The extreme value is at (2,4)

Therefore When the bird is 2 feet away from the first fence post, it reaches its minimum height of 4 feet , Minimum at (2,4)

Therefore Option C and D is the correct answer.

To know more about Quadratic Function

brainly.com/question/5975436

#SPJ1

7 0

2 years ago