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3 years ago

7

Let $\overline{AB}$ and $\overline{CD}$ be chords of a circle, that meet at point $Q$ inside the circle. If $AQ = 6,$ $BQ = 12$,

and $CD = 38$, then find the minimum length of $CQ$.

1 answer:

Cloud [144]3 years ago

3 0

9514 1404 393

Answer:

2

Step-by-step explanation:

The products of chord lengths are the same for the intersecting chords:

AQ×BQ = CQ×DQ

6×12 = CQ×(38 -CQ)

This gives a quadratic in CQ:

CQ² -38CQ +72 = 0 . . . . . write in standard form

(CQ -2)(CQ -36) = 0 . . . . . factor the quadratic

CQ = 2 or 36 . . . . . . . values of CQ that make the factors zero

The minimum length of CQ is 2 units. (DQ will be 36.)

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NNADVOKAT [17]

To solve this, you need to make the amount they built on Monday and the amount they built on Tuesday compatible with each other so you can add them together to subtract that amount from 4.

Start by multiplying 1/2 by 3, and 1/3 by 2.

This gives you 3/6 and 2/6. Now add them together:

3/6 + 2/6 = 5/6

Finally, subtract this amount from 4:

4 - 5/6 = 3 1/6 (19/6 works too)

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

Find the modulus of the complex number 6-2i

Papessa [141]

Answer:

The modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

Step-by-step explanation:

Given the number

$6-2i$

We know that

$z = x + iy$

where x and y are real and $\sqrt{-1}=i$

The modulus or absolute value of z is:

$|z|\:=\sqrt{x^2+y^2}$

Therefore, the modulus of $6-2i$ will be:

$z=6-2i$

$z=6+(-2)i$

$|z|\:=\sqrt{x^2+y^2}$

$|z|\:=\sqrt{6^2+\left(-2\right)^2}$

$=\sqrt{6^2+2^2}$

$=\sqrt{36+4}$

$=\sqrt{40}$

$=\sqrt{2^2}\sqrt{2\cdot \:5}$

$=2\sqrt{2\cdot \:5}$

$=2\sqrt{10}$

Therefore, the modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

3 0

2 years ago

PLEASEE HEEELP! In the normal distribution, 68% of the data lies within 1 standard deviation A: __/6 of the mean, 95% of the dat

Lena [83]

Answer:

a) 2.5% b) 50%

Step-by-step explanation:

1300 is two standard deviations higher than the mean. Since 95% of the data is covered within two standard deviations to the left and right of the mean, 5% is not covered. So, we have 2.5% leftover on the left side of the curve, under 900, and 2.5% leftover on the right side of the graph that is above 1300.

The mean is 1100, so anything above or below the mean is exactly 50% in a normal distribution.

3 0

3 years ago

Read 2 more answers

Plz help , due in 5 mins

Grace [21]

Answer:

option a) 30 : 19

Step-by-step explanation:

$Length = 7 \frac{1}{2} = \frac{15}{2}\\\\Width = 4\frac{3}{4} = \frac{19}{4}$

<em><u>Ratio of length to width:</u></em>

$\frac{15}{2} : \frac{19}{4}$ <em><u></u></em>

$=\frac{\frac{15}{2}}{\frac{19}{4}}\\\\=\frac{15}{2 } \times \frac{4}{19}\\\\=15\times \frac{2}{19}\\\\=\frac{30}{19}\\\\30 : 19$

3 0

3 years ago

Find the f^-1(x) and it’s domain

borishaifa [10]

Answer:

$f^{-1}(x) = (x + 8)^2$

$x \ge -8$

Step-by-step explanation:

Given

$f(x) = \sqrt x - 8$

Solving (a): $f^{-1}(x)$

We have:

$f(x) = \sqrt x - 8$

Express f(x) as y

$y = \sqrt x - 8$

Swap x and y

$x = \sqrt y - 8$

Add 8 to $both\ sides$

$x + 8 = \sqrt y - 8 + 8$

$x + 8 = \sqrt y$

Square both sides

$(x + 8)^2 = y$

Rewrite as:

$y = (x + 8)^2$

Express y as: $f^{-1}(x)$

$f^{-1}(x) = (x + 8)^2$

To determine the domain, we have:

The original function is $f(x) = \sqrt x - 8$

The range of this is: $f(x) \ge -8$

The $domain$ of the $inverse$ function is the $range$ of the $original$ function.

<em>Hence, the domain is:</em>

$x \ge -8$

3 0

3 years ago