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madreJ [45]
3 years ago
13

Divide 2x2 + 7x – 3 by 2x + 5. Which expression represents the quotient and remainder?

Mathematics
1 answer:
jeka57 [31]3 years ago
4 0

         ___x_ + 6______
2x - 5 | 2x^2  +  7x - 3
          - (2x^2 - 5x)
         -----------------------
                     12x   - 3
                   -(12x  - 30)
                 ------------------
                               27
Quotient    x + 6
Remainder 27  OR    27/(2x + 5)
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Which of the following constants can be added to x2 - 3x to form a perfect square trinomail?
serious [3.7K]

The constant that can be added to x^2 - 3x to form a perfect square trinomial is 2\frac{1}{4}

The given expression is

x^2 - 3x

To form a perfect square trinomial

(a-b)^2 =a^2-2ab+b^2

The given expression is

x^2 - 3x

first we have to add a constant term with it

x^2 - 3x + z

By comparing the given expression and the perfect square trinomial

a^2 =x^2

a = x

Similarly

-2ab = 3x

where know a =x

Then,

-2b = 3

b = -3/2

Similarly

b^2 = z

(\frac{-3}{2})^2 = z

9/4 = z

Convert the simple fraction to mixed fraction

9/4 = 2\frac{1}{4}

Hence, the constant that can be added to x^2 - 3x to form a perfect square trinomial is 2\frac{1}{4}

The complete question is :

Which of the following constants can be added to x2 - 3x to form a perfect square trinomial?

1\frac{1}{2},2\frac{1}{4} and 4\frac{1}{2}

Learn more about perfect square trinomial here

brainly.com/question/16615974

#SPJ1

4 0
1 year ago
If $1995.00 is shared equally among 5 men, how much would each get?
Vesna [10]

Answer:

each would receive $399

Step-by-step explanation:

divide 5 from 1995 which equals 399 :)

8 0
3 years ago
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Upper bound and lower bound of 17
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Answer:

Step-by-step explanation:

Hey there this si the upper and lower bound for 17

16.5<17<17.5

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Find the area of the shaded region. Round your answer to the nearest tenth.
Alex
Check the picture below on the left-side.

we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.

now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.

so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.

\bf \textit{area of a sector of a circle}\\\\&#10;A_x=\cfrac{\theta \pi r^2}{360}\quad &#10;\begin{cases}&#10;r=radius\\&#10;\theta =angle~in\\&#10;\qquad degrees\\&#10;------\\&#10;r=6\\&#10;\theta =60&#10;\end{cases}\implies A_x=\cfrac{60\cdot \pi \cdot 6^2}{360}\implies \boxed{A_x=6\pi} \\\\&#10;-------------------------------\\\\

\bf \textit{area of a segment of a circle}\\\\&#10;A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta )  \right]&#10;\begin{cases}&#10;r=radius\\&#10;\theta =angle~in\\&#10;\qquad degrees\\&#10;------\\&#10;r=6\\&#10;\theta =120&#10;\end{cases}

\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o )  \right]&#10;\\\\\\&#10;A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\&#10;-------------------------------\\\\&#10;\textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}

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4 years ago
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