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dedylja [7]
2 years ago
9

I am not sure on how to do this ​

Mathematics
2 answers:
sweet [91]2 years ago
8 0

Answer:

CP = 8 cm

Step-by-step explanation:

Given two chords that intersect inside a circle, then then the product of the parts of one chord is equal to the product of the parts of the other chord, that is

DP × CP = AP × PB ← substitute values

6 × CP = 12 × 4

6CP = 48 ( divide both sides by 6 )

CP = 8

olchik [2.2K]2 years ago
5 0

Answer:

8.

Step-by-step explanation:

Use the intersected chords of a circle  theorem:

AP * PB = CP * PD

12 * 4 =  CP * 6

6CP = 48

CP = 8.

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PLEASE HELP !
kirza4 [7]

Answer:

331.52 oz

Step-by-step explanation:

Google :D

6 0
2 years ago
Read 2 more answers
Let a and b be roots of x² - 4x + 2 = 0. find the value of a/b² +b/a²​
erastovalidia [21]

Answer:

\dfrac{a}{b^2}+\dfrac{b}{a^2}=10

Step-by-step explanation:

Given equation:   x^2-4x+2=0

The roots of the given quadratic equation are the values of x when y=0.

To find the roots, use the quadratic formula:

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Therefore:

a=1, \quad b=-4, \quad c=2

\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}

\textsf{Let }a=2+\sqrt{2}

\textsf{Let }b=2-\sqrt{2}

Therefore:

\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}

6 0
1 year ago
Read 2 more answers
Does (-15, -691) make the equation y = -51x - 74 true?<br><br><br>​
dimulka [17.4K]

Well, we can input both variables to check and see if the numbers hold.

-691 = -51(-15) - 74

-691 = 765-64

-691 = 701

(-15, -691) does not make the equation true.

3 0
3 years ago
Read 2 more answers
IfA=(-2,-4) and B=(-8,4), what is the length of AB?
lisabon 2012 [21]

Answer:

<h2>10 units </h2>

solution,

A(-2,-4)--->(X1,y1)

B(-8,4)----->(x2,y2)

now,

ab =  \sqrt{(x2 - x1) ^{2}  +  {(y2 - y1)}^{2} }  \\  \:  \:  \:  \:  =  \sqrt{ (- 8 - ( - 2)) ^{2} + (4 - ( - 4))  ^{2} }  \\  \:  \:  \:  =  \sqrt{( - 8 + 2) ^{2} +  {(4 + 4)}^{2}  }  \\  \:  \:  \:  =  \sqrt{( - 6) ^{2} +  {(8)}^{2}  }  \\  \:  \:  =   \sqrt{36 + 64}  \\  \:  \:  \:  =  \sqrt{100}  \\  \:  \:  \:  =  \sqrt{ {(10)}^{2} }  \\  \:  \:  \:  \:  = 10 \: units

Hope this helps..

Good luck on your assignment

3 0
2 years ago
The identifying letter at the top of a column is called the
Papessa [141]
Can you give more information on your question? 
8 0
3 years ago
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