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LOTS OF POINTS GIVING BRAINLIEST I NEED HELP PLEASEE

Sidana [21]

Answer:

Segment EF: y = -x + 8

Segment BC: y = -x + 2

Step-by-step explanation:

Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.

Upon observing the given diagram, we can infer the following corresponding sides:

$\displaystyle\mathsf{\overline{BC}\:\: and\:\:\overline{EF}}$

$\displaystyle\mathsf{\overline{BA}\:\: and\:\:\overline{ED}}$

$\displaystyle\mathsf{\overline{AC}\:\: and\:\:\overline{DF}}$

We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.

<h2>Slope of Segment BC:</h2>

In order to solve for the slope of segment BC, we can use the following slope formula:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}} }$

Use the following coordinates from the given diagram:

Point B: (x₁, y₁) = (-2, 4)

Point C: (x₂, y₂) = ( 1, 1 )

Substitute these values into the slope formula:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{1\:-\:4}{1\:-\:(-2)}\:=\:\frac{-3}{1\:+\:2}\:=\:\frac{-3}{3}\:=\:-1}$

<h2>Slope of Segment EF:</h2>

Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:

Point E: (x₁, y₁) = (4, 4)

Point F: (x₂, y₂) = (6, 2)

Substitute these values into the slope formula:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{2\:-\:4}{6\:-\:4}\:=\:\frac{-2}{2}\:=\:-1}$

Our calculations show that segment BC and EF have the same slope of -1. In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.

Since segments BC and EF have the same slope, then it means that $\displaystyle\mathsf{\overline{BC}\:\: | |\:\:\overline{EF}}$ .

<h2>Slope-intercept form:</h2><h3><u>Segment BC:</u></h3>

The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.

Using the slope of segment BC, m = -1, and the coordinates of point C, (1, 1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>

y = mx + b

1 = -1( 1 ) + b

1 = -1 + b

Add 1 to both sides to isolate b:

1 + 1 = -1 + 1 + b

2 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 2.

Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:

⇒ y = -x + 2.

<h3><u /></h3><h3><u>Segment EF:</u></h3>

Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>

y = mx + b

4 = -1( 4 ) + b

4 = -4 + b

Add 4 to both sides to isolate b:

4 + 4 = -4 + 4 + b

8 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 8.

Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:

⇒ y = -x + 8.

8 0

2 years ago

Solve four tenths plus four ninths

weeeeeb [17]

The answer is Eight ninetieths
Pls mark as brainliest thx

6 0

2 years ago

$700 is invested at 4% per annum. How long will it take for the amount to reach $784?

Blababa [14]

Answer:

3yrs

Step-by-step explanation:

Given parameters:

Capital = $700

Percentage interest = 4% per annum

Unknown:

Time it will take for amount to be = $784

Solution:

To find the simple interest on an amount;

I = $\frac{PRT}{100}$

P is the principal

R is the rate

T is the time

I is the interest

Insert the parameters:

I = $\frac{700x 4 x T}{100}$

I = 28T

So;

At the end of the duration:

Principal + Interest = Amount

700+28T = 784

28T = 784 - 700

28T = 84

T = 3yrs

5 0

2 years ago

Help within 20mins<br> Pls

Sergio039 [100]

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3 0

2 years ago

Sin(xy)-x=0 find dy/Dx

MariettaO [177]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2989024

——————————

You have y as an implicit function of x:

sin(xy) – x = 0

Use implicit differentiation. As y is a function of x, then you must apply the chain rule there:

d    d
—— [ sin(xy) – x ] = —— (0)
dx dx

d    d d
—— [ sin(xy) ] – —— (x) = —— (0)
dx    dx    dx

d
—— [ sin(xy) ] – 1  = 0
dx

d
—— [ sin(xy) ]  = 1
dx

d
cos(xy) · —— (xy) = 1
dx

Now, apply the product rule for that last derivative:

$\mathsf{cos(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\
\mathsf{cos(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\
\mathsf{y\,cos(xy)+x\,cos(xy)\cdot \dfrac{dy}{dx}=1}$

dy
Isolate —— :
dx

dy
x cos(xy) · —— = 1 – y cos(xy)
dx

Assuming x cos(xy) ≠ 0,

dy    1 – y cos(xy)
—— = ————————    <———    this is the answer.
dx    x cos (xy)

I hope this helps. =)

4 0

2 years ago