Ask question

Ask question

All categories

Norma-Jean [14]

3 years ago

6

Trim the exam the equation below and said that the equation has exactly one solution -8(x - 4) + 2x = -4(x - 8) - 2x which expla

ins whether Trina is correct:
a.) Trina is correct because the two sides of the equation are equivalent
b.) Trina is correct because the two sides of the equation are not equivalent
c.) Trina is not correct because the two sides of the equation are equivalent
d.) Trina is not correct because the two sides of the equation are not equivalent
I'm super confused on this can somebody help me

1 answer:

vivado [14]3 years ago

4 0

The answer is C Trina is not correct because the two sides of the equation are equivalent

You might be interested in

What is the equation of the perpendicular bisector of this line segment?

agasfer [191]

X=2

Explanation:
You need to find the midpoint of the segment

-2+6=4
4/2=2

3 0

3 years ago

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

<u />

(b) The rate the area formed by the ladder is changing is approximately <u>-75.29 ft.²/sec</u>

<u />

(c) The rate at which the angle formed with the wall is changing is approximately <u>0.286 rad/sec</u>.

Reasons:

The given parameter are;

Length of the ladder, <em>l</em> = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let <em>y</em> represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ <u>-6.86 m/s downwards</u>

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ <u>-75.29 ft.²/sec</u>

(c) From trigonometric ratios, we have;

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ <u>0.286 rad/sec</u>

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago

Help i need help fast i will give 5 stars <br> find 3/7 in 14

9966 [12]

Are you asking to find 3/7 of 14? In that case the answer would be 6. 14 divided by 7 is 2 and 2x3 is 6!

4 0

3 years ago

a form of reasoning called _ is the process of forming general ideas and rules based on your experiences and observation

Slav-nsk [51]

Answer:

what the question that your asking

6 0

3 years ago

If r1 , r2 , and r3 represent rotations from Dn and f 1 , f 2 , and f 3 represent reflections from Dn , determine whether r1 r2

ziro4ka [17]

Answer:

It is reflection

Step-by-step explanation:

We know these three rules:

-a rotation followed by a rotation is a rotation

-a reflection followed by a reflection is also a rotation

-a rotation folowed by a reflection is a reflection

So we have: f3r3 is f (reflection)

f2 f is r (reflection)

r3 r is r

f2 r is f

r2 f is f

r1 f is f

So, r1r2f1r3f2f3r3 is reflection.

7 0

3 years ago