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1 year ago

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(Please help )Prove or disprove that point (4.00, 4,50) lies on circle O centered at the origin.

1 answer:

Marat540 [252]1 year ago

4 0

Answer:

go ahead and view TaraLM she's making free credit reports on her phone she's on the contact list this is

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Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago

crimeas [40]

Answer:

The zeros are:

$x =4, x=2, x = 5$

The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

$h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)$

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

$0=\left(x-4\right)^2\left(x^2-7x+10\right)$

switch sides

$\left(x-4\right)^2\left(x^2-7x+10\right)=0$

as

$x^2-7x+\:10=\left(x-2\right)\left(x-5\right)$

so

$\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

$\mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

so

$x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x =4, x=2, x = 5$

Thus, the zeros are:

$x =4, x=2, x = 5$

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

The function has three distinct real zeros.

Hence, option (B) is true.

4 0

2 years ago

Determine whether each question is biased. Explain your answer. please help!!

LenKa [72]

I’m having a hard time understanding your question

3 0

2 years ago

Write as a product:<br><br> x^2+ax^2–y–ay+cx^2–cy<br> Can you plz help 20 points for correct answer.

lana66690 [7]

download an app called photomath

3 0

2 years ago

Drag the correct symbol into the box to compare expessions<br> 6^2-2^2___4(6+2)<br> A.≠<br> B.=

Mnenie [13.5K]

Simplify both sides to compare:

6^2 - 2^2 = 36-4 = 32

4(6+2) = 4 x 8 = 32

they are equal to each other.

Answer: B. =

8 0

1 year ago