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3 years ago

(2x^2+4x-3)-(2x^2+4x-3) show work mark brainliest

Mathematics

1 answer:

Aleksandr [31]3 years ago

8 0

$Solution: \left(2x^2+4x-3\right)-\left(2x^2+4x-3\right)=0$

Steps:

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a, =2x^2+4x-3-\left(2x^2+4x-3\right)$

$-\left(2x^2+4x-3\right):\quad -2x^2-4x+3$

$=2x^2+4x-3-2x^2-4x+3$

$\mathrm{Simplify}\:2x^2+4x-3-2x^2-4x+3:\quad 0$

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Given quadrilateral RSTU, determine if each pair of sides (if any) are parallel and which are perpendicular for the

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Answer:

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Basically anywhere that there's lines that make right angles are perpendicular to each other

I hope this helps

Step-by-step explanation:

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3 years ago

Two spherical balloons are filled with water. The first balloon has a radius of 3 cm, and the second has a radius of 6 cm.

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3 years ago

The maximum value of 12 sin 0-9 sin²0 is: -

jeka57 [31]

Answer:

Step-by-step explanation:

The question is not clear. You have indicated the original function as 12sin(0) - 9sin²(0)

If so, the solution is trivial. At 0, sin(0) is 0 so the solution is 0

However, I will assume you meant the angle to be $\theta$ rather than 0 which makes sense and proceed accordingly

We can find the maximum or minimum of any function by finding the first derivate and setting it equal to 0

The original function is

$f(\theta) = 12sin(\theta) - 9 sin^2(\theta)$

Taking the first derivative of this with respect to $\theta$ and setting it equal to 0 lets us solve for the maximum (or minimum) value

The first derivative of $f(\theta)$ w.r.t $\theta$ is

$12\cos\left(\theta\right)-18\cos\left(\theta\right)\sin\left(\theta\right)$

And setting this = 0 gives

$12\cos\left(\theta\right)-18\cos\left(\theta\right)\sin\left(\theta\right) = 0$

Eliminating $cos(\theta)$ on both sides and solving for $sin(\theta)$ gives us

$sin(\theta) = \frac{12}{18} = \frac{2}{3}$

Plugging this value of $sin(\theta)$ into the original equation gives us

$12(\frac{2}{3}) - 9(\frac{4}{9} ) = 8 - 4 = 4$

This is the maximum value that the function can acquire. The attached graph shows this as correct

3 0

2 years ago

Please help!!! I’ll mark as brainliest :)

lyudmila [28]

Answer:

the third one

Step-by-step explanation:

5 0

3 years ago

PLEASE HELP IM ON A TIME LIMIT :(

Mice21 [21]

Answer:

hey hope this helps

<h3 /><h3>Comparing sides AB and DE </h3>

AB =

$\sqrt{ {1}^{2} + {1}^{2} }$

$= \sqrt{2}$

$= \sqrt{ {(3 - 5)}^{2} + {(1 + 1}^{2} } \\ = \sqrt{ {( - 2)}^{2} + {(2)}^{2} } \\ = \sqrt{4 + 4} \\ = \sqrt{8} \\ = 2 \sqrt{2}$

So DE = 2 × AB

and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.

scale factor = AB/DE

= 2

It's been reflected across the Y-axis

moved thru the translation of 3 units towards the right of positive x- axis 

for this let's compare the location of points B and D

For both the y coordinate is same while the x coordinate of B is 0 and that of D is 3

so the triangle has been shifted by 3 units across the positive x axis

7 0

2 years ago