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3 years ago

Which statement is true about the equation fraction 3 over 4z − fraction 1 over 4z + 3 = fraction 2 over 4z + 5?

Mathematics

2 answers:

alexira [117]3 years ago

6 0

I believe it has infinite solutions? Not sure

You can graph this via graphing calculator and see that the lines intersect almost infinitely.

Allisa [31]3 years ago

3 0

Answer:

TOOK THE TEST ITS A. NO SOLUTION

Step-by-step explanation:

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Which term can be added to the list so that the greatest common factor of the three terms is 1217

kifflom [539]

Answer:

1217

Step-by-step explanation:

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3 years ago

I need to know the answer to question 10

motikmotik

The answer is actually e although it would have been a

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3 years ago

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timofeeve [1]

Thes answer would be 51 or -51

6 0

3 years ago

Read 2 more answers

Which triangle is △MNO similar to and why? △MNO is similar to △GHK by AA Similarity Postulate . △MNO is similar to △DEF b

andrew11 [14]

<span>MNO is similar to GHK by AA Similarity Postulate

Let's start by listing each triangle and the measurements of all three angles. For each triangle, we've been given the measurements of 2 of the angles and the 3 angle will simply be 180 minus the other 2 angles. I assume you can do the subtraction, so I'll simply list each triangle with all three angle measurements.
NMO: 79, 22, 79
GHK: 79, 79, 22
PQR: 20, 79, 81
DEF: 82, 22, 76
And the triangles NMO and GHK are similar to each other since they have the same angles. The order really doesn't matter since it's OK for similar triangles to be rotated or reflected. The key thing to remember in a triangle is that if you've been told what 2 of the angles are, you also know what the 3rd angle is since the sum of the angles of a triangle will always be 180.
So the answer is:
MNO is similar to GHK by AA Similarity Postulate"</span>

7 0

4 years ago

Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2<=1}.

castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$f_x = 6x^2, f_y=4y^3$

Hence, $f_x = f_y = 0$ if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$x^2+y^2=1$

Hence, the top half of the boundary is,

$$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}$

On T we have, $f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1 \text{ for}\ -1 \leq x \leq 1$

We compute

$\frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and only if x=0, x= 1/2 or x = -2.

We disregard $x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $$( 1/2, -\sqrt3/2)$

It remains to evaluate f(x, y) at the points $(0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\ (\pm1, 0)$ .

We should consider latter two points, $(\pm1,0)$ , since they are the boundary points for the T and also B. We compute $f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2$

We conclude that the absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0

3 years ago