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GaryK [48]
3 years ago
13

3+4 40+3 30+4 which of these is another way to write

Mathematics
2 answers:
zhuklara [117]3 years ago
8 0
(3+4) (40+3) (30+4) 
= 7 + 47 + 34
= (7+47) +35
=  54 + 35 
=   89
Andre45 [30]3 years ago
8 0
30-4 30+4 4+30 I think
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Write 3x^2-18x-6 in vertex form
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The standard form of a quadratic equation is \displaystyle{ y=ax^2+bx+c, while the vertex form is:

                      y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:

\displaystyle{ y=3(x^2-6x-2).


Second, we notice that x^2-6x are the terms produced by (x-3)^2=x^2-6x+9, without the 9. So we can write:

x^2-6x=(x-3)^2-9, and substituting in \displaystyle{ y=3(x^2-6x-2) we have:

\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

Finally, distributing 3 over the two terms in the brackets we have:

y=3[x-3]^2-33.


Answer: y=3(x-3)^2-33
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3 years ago
Which fraction has a terminating decimal as its decimal expansion ?
Alekssandra [29.7K]

Answer:

b

Step-by-step explanation:

1/5 is equal to 20% =.2 it stops there

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2 years ago
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You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o
Aleksandr [31]

Answer:

Therefore the circumference of the circle is =\frac{20\pi}{4+\pi}

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

\Rightarrow s=\frac{10-\pi r}{2}

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

A=\pi r^2+ (\frac{10-\pi r}{2})^2

\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2

\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}

\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25

For maximum or minimum \frac{dA}{dr}=0

Differentiating with respect to r

\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi

Again differentiating with respect to r

\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}    > 0

For maximum or minimum

\frac{dA}{dr}=0

\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0

\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}

\Rightarrow r=\frac{10}{4+\pi}

\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0

Therefore at r=\frac{10}{4+\pi}  , A is minimum.

Therefore the circumference of the circle is

=2 \pi \frac{10}{4+\pi}

=\frac{20\pi}{4+\pi}

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