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Katyanochek1 [597]

3 years ago

9

In circle O below, AB is a diameter and m

1 answer:

Neko [114]3 years ago

4 0

Answer:

i don't know

Step-by-step explanation:

i seriously don't know

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What are the zeros for each function ? f(x) = 2(x+1)(x-3)(x-1) ^2 (squared)

Setler79 [48]

If you do 5*the ratio of ur mom then u will get Y which times ur dads ding dong get u f(x)= 758984327934792379427893.

4 0

2 years ago

for what value of k, the line joining 3x-ky+7=0 is perpendicular to the line joining (4 ,3) and ( 5, -3).

Schach [20]

Answer:

k = 18

=========

<h2>Given</h2>

<h3>Line 1</h3>

3x - ky + 7 = 0

<h3>Line 2</h3>

Passing through the points (4, 3) and (5, - 3)

<h2>To find</h2>

The value of k, if the lines are perpendicular

<h2>Solution</h2>

We know the perpendicular lines have opposite reciprocal slopes, that is the product of their slopes is - 1.

Find the slope of line 1 by converting the equation into slope-intercept from standard form:

<u><em>Info:</em></u>

<em>standard form is ⇒ ax + by + c = 0, </em>
<em>slope - intercept form is ⇒ y = mx + b, where m is the slope</em>

3x - ky + 7 = 0
ky = 3x + 7
y = (3/k)x + 7/k

Its slope is 3/k.

Find the slope of line 2, using the slope formula:

m = (y₂ - y₁)/(x₂ - x₁) = (-3 - 3)/(5 - 4) = - 6/1 = - 6

We have both the slopes now. Find their product:

(3/k)*(- 6) = - 1
- 18/k = - 1
k = 18

So when k is 18, the lines are perpendicular.

4 0

2 years ago

Need help ASAP please

Sveta_85 [38]

Answer:

bruh ur so dumb

Step-by-step explanation:

google it

5 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

4 years ago

What does this equal? 3(x+5)=7-x

True [87]

3(x+5)=7-x3x+15=7-x4x=-8x=-2

3 0

3 years ago

Read 2 more answers