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kompoz [17]
3 years ago
7

The perimeter of a square is represented by the expression 32x - 12.8 find the side length of the square.

Mathematics
1 answer:
djverab [1.8K]3 years ago
8 0

Given:

Perimeter of a square = 32x-12.8

To find:

The side length of the square.

Solution:

We know that, the perimeter of the square is 4 times of its side length.

P=4a

Where, a is the side length.

Perimeter of a square = 32x-12.8

Now,

4a=32x-12.8

Divide both sides by 4.

a=\dfrac{32x-12.8}{4}

a=\dfrac{32x}{4}-\dfrac{12.8}{4}

a=8x-3.2

Therefore, the side length of the square is 8x-3.2 units.

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Desirable outcomes = TTH, THT, TTH   (3 possibilities)

Required probability = 3/8
6 0
3 years ago
Read 2 more answers
The minimum/maximum value of the function y = a(x − 2)(x − 1) occurs at x = d, what is the value of d?
daser333 [38]

Answer:

d=\frac{3}{2}=1.5

Step-by-step explanation:

We have the function:

y=a(x-2)(x-1)

And we want to find x=d for which the minimum/maximum value will occur.

Notice that our function is a quadratic in factored form.

Remember that the minimum/maximum value always occurs at the vertex point.

And remember that the x-coordinate of the vertex is the axis of symmetry.

Since a quadratic is always symmetrical on both sides of its axis of symmetry, a quadratic’s axis of symmetry is the average of the two roots/zeros of the quadratic.

Therefore, the value x=d such that it produces the minimum/maximum value is the average of the two roots.

Our factors are <em>(x-2) </em>and <em>(x-1)</em>.

Therefore, our roots/zeros are <em>x=1, 2</em>.

So, the average of them are:

d=\frac{1+2}{2}=3/2=1.5

Therefore, regardless of the value of <em>a</em>, the minimum/maximum value will occur at <em>x=d=1.5</em>.

Alternative Method:

Of course, we can also expand to confirm our answer. So:

y=a(x^2-2x-x+2)\\y=a(x^2-3x+2)\\y=ax^2-3ax+2a

The x-coordinate of the vertex is still going to be the place where the minimum/maximum is going to occur.

And the formula for the vertex is:

x=-\frac{b}{2a}

So, we will substitute <em>-3a</em> for <em>b</em> and <em>a</em> for <em>a</em>. This yields:

x=-\frac{-3a}{2a}=\frac{3}{2}=1.5

Confirming our answer.

4 0
3 years ago
What is ? <br> algebra 2
Alexxx [7]

(?) = (x - 3)(x + 3)

Step-by-step explanation:

[Attached]

4 0
3 years ago
Which of the following correctly expresses sin (70) + sin(30) as a product?
JulsSmile [24]

Answer:

2sin50 cos20

Step-by-step explanation:

We need to write sin (70) + sin(30) as a product. The formula used here is :

\sin A+\sin B=2\sin (\dfrac{A+B}{2})\cos(\dfrac{a-b}{2})

Here, A = 70 and B = 30

So,

\sin 70+\sin 30=2\sin (\dfrac{70+30}{2})\cos(\dfrac{70-30}{2})\\\\\sin 70+\sin 30=2\sin 50\cos20

So, the value of sin (70) + sin(30) is 2sin50 cos20. Hence, the correct option is  (c).

7 0
3 years ago
Solve x^3-7x^2+7x+15​
ruslelena [56]

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

6 0
3 years ago
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