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3 years ago

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Want money answer this.

2 answers:

jok3333 [9.3K]3 years ago

8 0

Answer: CM:ME (6.71:4.47) = 1.5 : 1

Step-by-step explanation:

the coordinates of the endpoints of CE are C(-3, -7) and E(7, -2). point M (3, -4) is on CE. What is the ratio of CM:ME? This is a distance formula problem, where you are comparing the two ratio's of CM to ME.

It is well worth your time to memorize the distance formula: d = sqrt ((x - x1)2 + (y - y1)2) using two points: (x,y), (x1,y1). You will use this many, many times in your math classes.

CM: sqrt ((-3 - 3)2 + (-7 - -4)2) = sqrt ((6)2 + (-3)2) = sqrt (45) ≅ 6.71

ME: sqrt ((3 - 7)2 + (-4 - -2)2) = sqrt ((-4)2 + (-2)2) = sqrt (20) ≅ 4.47

CM:ME (6.71:4.47) = 1.5 : 1

Lorico [155]3 years ago

6 0

Answer:

Is this a scam

Step-by-step explanation:

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BaLLatris [955]

Answer:

Step-by-step explanation:

5c+1+3c=61

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4 0

3 years ago

Find the exact value of csc theta if tan theta = sqrt3 and the terminal side of theta is in Quadrant III.

WARRIOR [948]

Answer:

3rd option

Step-by-step explanation:

Using the identities

cot x = $\frac{1}{tanx}$

csc² x = 1 + cot² x

Given

tanθ = $\sqrt{3}$ , then cotθ = $\frac{1}{\sqrt{3} }$

csc²θ = 1 + ( $\frac{1}{\sqrt{3} }$ )² = 1 + $\frac{1}{3}$ = $\frac{4}{3}$

cscθ = ± $\sqrt{\frac{4}{3} }$ = ± $\frac{2}{\sqrt{3} }$

Since θ is in 3rd quadrant, then cscθ < 0

cscθ = - $\frac{2}{\sqrt{3} }$ × $\frac{\sqrt{3} }{\sqrt{3} }$ = - $\frac{2\sqrt{3} }{3}$

8 0

3 years ago

Given f(x) = 10-2x find f(7)

lara [203]

10 - 2x
10 - 2(7)
= 10 - 14
= -4

8 0

3 years ago

Read 2 more answers

If 0°<0<90° and sin0=12/13, find cos0 using trigonometric identities.

Aleks04 [339]

here's the solution,

we know :

$\sin( \theta) = \dfrac{perpendicular}{hypotenuse}$

So,

$\dfrac{p}{h} = \dfrac{12}{13}$

so.. let the perpendicular be 12x and hypotenuse be 13x

now,

by applying pythagoras theorem,

$b {}^{2} = h {}^{2} - p {}^{2}$

where,

b = base
h = hypotenuse
p = perpendicular

So,

$b {}^{2} = ({13x})^{2} - ({12x})^{2}$
$b {}^{2} = 169x {}^{2} - 144x {}^{2}$
${b}^{2} = 25x {}^{2}$
$b = 5x$

so,

$\cos( \theta) = \dfrac{b}{h}$

$\cos( \theta) = \dfrac{5x}{13x}$

$\cos( \theta) = \dfrac{5}{13}$

hope it helps !!

3 0

3 years ago

Helppppppp i need help

Lunna [17]

Do you need the fraction? I don’t understand what you are asking

8 0

3 years ago