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3 years ago

Find the slope of the line contain the given pair of points (-6,4) and (1,0)

2 answers:

8 0

$(-6,4) \\
x_1=-6 \\
y_1=4 \\ \\
(1,0) \\
x_2=1 \\
y_2=0 \\ \\
\hbox{the slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{0-4}{1-(-6)}=\frac{-4}{1+6}=-\frac{4}{7}$

The slope is -4/7.

gulaghasi [49]3 years ago

8 0

$m = \frac{ y_{2} - y_{1}}{ x_{2} - x_{1}}
 







$
$m = \frac{0 - 4}{1 - (-6)}$
$m = \frac{0 - 4}{1 + 6}$
$m = \frac{-4}{7}$

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What are the coordinates of the x-intercepts of the parabola y = x² - 8x + 15?

tamaranim1 [39]

Answer:

(3, 0) and (5, 0)

Step-by-step explanation:

we have

$y=x^{2}-8x+15$

we know that

The x-intercepts are the values of x when the value of y is equal to zero

For y=0

$x^{2}-8x+15=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}-8x+15=0$

$a=1\\b=-8\\c=15$

substitute in the formula

$x=\frac{-(-8)\pm\sqrt{-8^{2}-4(1)(15)}} {2(1)}$

$x=\frac{8\pm\sqrt{4}} {2}$

$x=\frac{8\pm2} {2}$

$x=\frac{8+2} {2}=5$

$x=\frac{8-2} {2}=3$

x=3, x=5

therefore

The x-intercepts are (3,0) and (5,0)

4 0

3 years ago

A cereal box has a height of 32 cm. The base has 160 square cm what is the volume in cubic cm of the box

ruslelena [56]

Hello There!

All you really need to do is multiply 160 with 32.

$160$ × $32$ ≈ $5120$

5120 cm

Hope this helped!

7 0

3 years ago

Read 2 more answers

A.645,473<br> B.2,151,575<br> C.8,175,987<br> D.10,542,719

ELEN [110]

21515754 people total* 0.38= 8175986.52 people

Final answer: C

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3 years ago

One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides

Elina [12.6K]

<h2>Answer:</h2>

<em> The side of the triangle is either 38.63ft or 10.35ft</em>

<h2>Step-by-step explanation:</h2>

This problem can be translated as an image as shown in the Figure below. We know that:

The side of the square is 10 ft.
One of the vertices of an equilateral triangle is on the vertex of a square.
Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

$\mathbf{(1)} \ x^2=100+y^2$

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

$y+(10-y)=10$

Therefore, for Triangle 3, we have that by Pythagorean theorem:

$(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2$

Matching equations (1) and (2):

$2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0$

Using quadratic formula:

$y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68$

Finding x from (1):

$x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft$

<em>Finally, the side of the triangle is either 38.63ft or 10.35ft</em>

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3 years ago

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krek1111 [17]

Answer:

For 90:

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Step-by-step explanation:

6 0

3 years ago

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