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stira [4]
3 years ago
5

A rectangular school banner has a length of 60 inches and a width of 48 inches. A sign is made that is similar to the school ban

ner and has a length of 14 inches. What is the ratio of the area of the school banner to the area of the sign?
Mathematics
1 answer:
natulia [17]3 years ago
3 0

Answer:

The ratio of the length of the banner to the length of the sign = 54/17

So....since they are similiar figures, the widths will have the same ratio

And the ratio of the area of the banner to the area of the sign will be the square of this ratio  

Step-by-step explanation:

[54/17]^2  = 2916/289

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please help:) thank youuuyy
Lana71 [14]

9514 1404 393

Answer:

  • x = 8
  • BF = 45

Step-by-step explanation:

Using the given information, we have ...

  BH = 2·HF

  3x +6 = 2(2x -1)

  3x +6 = 4x -2

  8 = x

__

The length of the median is ...

  BF = BH + HF

  BF = (3x +6) +(2x -1) = 5x +5

  BF = 5·8 +5

  BF = 45

6 0
2 years ago
Find the domain of the function (13/x-11)
kotykmax [81]
The domain of 13/x - 11 is all real numbers except x cannot equal zero
8 0
2 years ago
Please help me, I will give brainliest
ipn [44]

Answer:

        8w² < 4w(150-w)

Step-by-step explanation:

Square area of  living  : w  · w  =  w²

Money spent : 8 · w²

Square area of  artichokes : (150 - w) · w    

Money earned : 4 · w · (150 - w)

Julia manages to save some money every week. That means that the money earned is bigger than the money spent ( the money spent is less than the money earned)

8 · w² < 4 · w · (150 - w)

7 0
2 years ago
The midpoint of gh is m(4,-3). One endpoint is G(-2,2). Find the coordinates of endpoint H
Mars2501 [29]

Answer:

Step-by-step explanation:

(-2 + x)/2= 4

-2 + x = 8

x = 10

(2 + y)/2 = -3

2 + y = -6

y = -8

(10, -8) for endpoint H

4 0
2 years ago
Question 3 (4 MARKS) Find the equation of the tangents to the curve of the function with rule y = x^2 at the points where x = a
Arte-miy333 [17]

Answer:

(0 , -a²)

Step-by-step explanation:

tangent at x = a and x = -a

y = x²     (a , a²) and (-a , a²) must be on the curve and tangent to curve

gradient dy/dx = 2x

slope (m) at x = a is <u>2a</u>     and slope (m') at x=-a is<u> -2a</u>

line1: (y - a²) / (x - a) = 2a

        y - a² = 2a (x - a)                           y = 2ax - a²   ... (1)

line2: (y - a²) / (x + a) = - 2a                   y = -2ax- a²   ... (2)

(1) - (2): 4ax = 0       a≠0    x = 0

y = - a²

I wish I did it right, or ....

3 0
2 years ago
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