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3 years ago

I'd like some help, thanks!

Mathematics

2 answers:

Otrada [13]3 years ago

7 0

It is likely D I am not positive

ANTONII [103]3 years ago

5 0

we know,

angle b= angle c

=>AB=AC(lines opposite to equal angles are equal)

=> 1/2x+5=5/6x+7

=>5/6x-1/2x= 5-7

=>2x/6= -2

=>x= -6

Now,

BC= 2/3x+7

=2/3×-6 +7

=-4 +7 = 3

therefore BC=3

I hope this is correct and helps you

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Simplify the expression:z2-z(z+3)+3z

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3 years ago

The probability of a certain plant growing taller than 4 meters is 1/3. If I plant 2 of these plants, find the probability that

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$\frac{1}{3}^2\\\frac{1}{3}*\frac{1}{3}\\\frac{1}{9}$

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3 years ago

Can someone help me with this

ludmilkaskok [199]

Answer:

Step-by-step explanation:

1/2r +1/2= -2/7r -29/7 Subtracting 1/2 from both sides

1/2r +1/2 - 1/2= -2/7r -29/7- 1/2

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2 years ago

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second

I'd like some help, thanks!​

I'd like some help, thanks!