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3 years ago

7

you are picking up a friend and going to school your friend lives 12 miles west of your house in school is 1.5 miles east of you

r friends when you are at school how far are you away from your home are you

1 answer:

IceJOKER [234]3 years ago

8 0

I think that your school is 10.5 miles away from your house

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Evaluate 3x +2y - 4z/x for x =5, y=-2,z=10

Evgen [1.6K]

3x + 2y - 4z/x for x=5, y=-2, z=10

3x + 2y - 4z/x

3(5) + 2(-2) - 4(10)/(5)

15 - 4 - 40/5

15 - 4 - 8

11 - 8

3

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3 years ago

The length of the rectangle is six times its width if the perimeter is 70cm find its area

Vesna [10]

Answer:

Step-by-step explanation:

Perimeter is 70 m

so w + w + 6w + 6w = 70

w = 70/14 = 5, l = 30m.

Area = l * w = 30 * 5 = 150 sq m

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3 years ago

Given an equilateral triangle that has an area of 1959 cm2. What is the length of its sides?

Nina [5.8K]

Equilateral
equi means equal
lateral means sides

all sides are equal
triangle has 3 sides

perimiter=side1+side2+side3
side1=side2=side3

area=1/2 times base times height
base=side
draw a line down the middle of te tiangle so you get 2 right triangles side by side
1/2base is one leg
base is hypotonuse
height is unknown
equilateral means that it is a 30-60-90 triangle with hypotonuse equal to 2x, height=x√3, bottom is x
1/2 times base times height=1959
1/2 times x times x√3=1959
multiply both sides by 2
(x^2)√3=3918
divide both sides by √3
x^2=3918/√3
square root both sides
x=47.5611
side=hypotonuse=2x
2x=2*47.5611
side=95.12222

legnth of each side is $2\sqrt{ \frac{3918}{ \sqrt{3} } }$ cm or about 95.12222 cm

7 0

3 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.

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3 years ago

Alan is giving a basic math test in his class. One of the questions in the test is about finding two factors of the number 221.

NARA [144]

Answer:

★ The two factors of the number 221 are 13 and 17.

A factor refers to a whole number, which fits evenly into another whole number. Thus, factors are the numbers, which we multiply together to get another number. The two numbers that can be multiply together to get 221 are 13 and 17.

Step-by-step explanation:

Hope you have a great day :)

6 0

3 years ago