Does the relationship show a ratio or a rate?

PLS I NEED HELP ANSWER WILL GIVE BRAINLIEST ANSWER BOTH QUESTIONS ONE AND TWO

KiRa [710]

I hope this is helpful

8 0

3 years ago

A rug maker is using a patter that is a rectangle with a length of 96 inches and a width of 60 inches. the rug maker wants to in

dexar [7]

Recall that the formula for the perimeter of a rectangle is: 2Length+2Width
With this in mind, we have that the original design is 96×60, where 90 is length and 60 is width, so it would simply be:
2(96+l) + 2 (60+w) = New Perimeter
Or simplified, it'd be:
2l + 2w + 312 = New Perimeter

8 0

3 years ago

one third of a land is on maize, a fifth of the remainder is on potatoes and the rest which is 0.5 ha is on beans. how much of t

disa [49]

Answer: $1250\ m^2$

Step-by-step explanation:

Let x= Total land.

Then for Maize = $\dfrac{x}{3}$

For Potatoes = $\dfrac15(x-\dfrac {x}{3})=\dfrac15(\dfrac{2x}{3})=\dfrac{2}{15}x$

For beans = $x- (\dfrac{x}{3}+\dfrac{2}{15}x)=0.5 \ ha$

$\Rightarrow\ x- (\dfrac{5x+2x}{15})=0.5 \ ha\\\\\Rightarrow\ x- (\dfrac{7x}{15})=0.5 \ ha\\\\\Rightarrow\ \dfrac{15x-7x}{15}=0.5 \ ha\\\\\Rightarrow\ \dfrac{8x}{15}=0.5 \ ha\\\\\Rightarrow\ x=0.5\times\dfrac{15}{8} \ ha\\\\\Rightarrow\ x=0.9375 \ ha$

1 ha = 10000.00 m²

0.9375 ha = 9375 m²

Land for potatoes = $\dfrac{2}{15}(9375) =1250\ m^2$

Hence, $1250\ m^2$ of the land is on potatoes.

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.

8 0

3 years ago

If there was a 42 degree climb of 32 degrees what would the equation of degrees be?

Anon25 [30]

Answer: Each degree is divided into 60 minutes as usual, but there are no seconds. The minute is expressed as a decimal instead. For example 23° 34.62' is read as "23 degrees 34.62 minutes. See also Nautical Angle Calculator

Step-by-step explanation:

also the story is maxed so do you won't to do it here when your done with your work.

7 0

3 years ago