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2 years ago

calculate the mid points of the line segments below using the midpoint formula given two endpoints (9,7) (-8, 10)

Mathematics

1 answer:

12345 [234]2 years ago

6 0

Answer: (0.5, 8.5)

Step-by-step explanation: (9+-8/2) is 1/2 and (7 + 10/ 2) is 8.5. After plugging it into the equation, I checked with a graphing site.

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nasty-shy [4]

Answer: yay thxs

Step-by-step explanation:

5 0

2 years ago

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

5 0

2 years ago

Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes

Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11 .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2) ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10 ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5 ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0 "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10 .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5 .............(8)

So orthocentre is at (37/10, 19/5) as in part A.

8 0

3 years ago

A circle has a radius of 5. An arc of 162 degrees. What is the length of the arc?

Hunter-Best [27]

Answer:

242

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Please help, easy geometry

iVinArrow [24]

Answer:

26 cm²

Step-by-step explanation:

The area of the rectangle whose dimensions are shown at the right and bottom is ...

(6 cm)(7 cm) = 42 cm²

The figure is smaller than that by the area of the space whose dimensions are shown at the right and in the middle left:

(4 cm)(4 cm) = 16 cm²

The figure area is then the difference ...

42 cm² - 16 cm² = 26 cm²

_____

<em>Alternate solution</em>

Draw a diagonal line between the lower right inside corner and the lower right outside corner. This divides the figure into two trapezoids.

The trapezoid at lower left has bases 7 and 4 cm, and height 6-4 = 2 cm. Its area is ...

A = (1/2)(b1 +b2)h = (1/2)(7 + 4)(2) = 11 . . . . cm²

The trapezoid at upper right has bases 6 cm and 4 cm and height 3 cm. Its area is ...

A = (1/2)(b1 +b2)h = (1/2)(6 + 4)(3) = 15 . . . . cm²

Then the area of the figure is the sum of the areas of these trapezoids, so is ...

11 cm² + 15 cm² = 26 cm²

_____

<em>Comment on other alternate solutions</em>

There are many other ways you can find the area of this figure. It can be divided into rectangles, triangles, or other figures of your choice. The appropriate area formulas should be used, and the resulting partial areas added or subtracted as required.

You can also let a geometry program find the area for you. (It is 26 cm².)

8 0

3 years ago

calculate the mid points of the line segments below using the midpoint formula given two endpoints (9,7) (-8, 10)​

calculate the mid points of the line segments below using the midpoint formula given two endpoints (9,7) (-8, 10)