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2 years ago

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:

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

I am not sure on how to do this ​

I am not sure on how to do this