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shutvik [7]
3 years ago
8

A map has a scale of 1 in. : 5 mi. The distance on the map between two cities is 11.5 inches. Find the actual distance between t

he cities
Mathematics
1 answer:
Vladimir [108]3 years ago
6 0

Answer:

57.5 mi

Step-by-step explanation:

1 in = 5 mi

11.5 × 5 = 57.5 mi

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Evaluate the expression for the given values.<br><br> 2x + (3y + z2)<br> X = 2, y = 3, z = 4
bazaltina [42]

Answer:

29

Step-by-step explanation:

2(2)+(3(3)+(4²))

4+(9+16)

4+25

29

pls if this was h helpful pls tag brainiest

3 0
2 years ago
If T(x, y) = (x + 5, y + 6) and Pris the image of P, what is the rule for the translation in which P is the image of P'? T(x, y)
xxTIMURxx [149]

Answer:

The translation is;

T(x,y)=(x-5,y-6)

Explanation:

Given the translation rule;

T(x,y)=(x+5,y+6)

when P' is the image of P.

For the inverse, when P is the image of P', the Translation rule would become;

\begin{gathered} x=x^{\prime}+5 \\ x^{\prime}=x-5 \\ y=y^{\prime}+6 \\ y^{\prime}=y-6 \\ So,\text{ the translation rule becomes;} \\ T(x,y)=(x-5,y-6) \end{gathered}

The translation is;

T(x,y)=(x-5,y-6)

3 0
1 year ago
The figures are similar. The ratio of the perimeters is 8 : 5. Find X
vova2212 [387]

Answer:

Two figures that have the same shape are said to be similar. When two figures are similar, the ratios of the lengths of their corresponding sides are equal. To determine if the triangles below are similar, compare their corresponding sides.

3 0
3 years ago
A six-sided number cube is rolled, and then a spinner with 5 equal sections labeled A through E is spun.
natka813 [3]
16 I think if not 130
4 0
2 years ago
Read 2 more answers
Is Part A correct? And can someone please help me with part B? (Highlighted)
Masja [62]
Part A. Your model is as good as any. It is hard to tell if the label needs to include the descriptor (length = 12 ft, for example). Certainly your model is sufficient for Part B.

Part B. The total area is the sum of the areas of each of the 6 faces of the box. Opposite faces are the same area, so you have
  A = 2(LW +WD +LD) = 2(LW +D(L +W))
Subsituting the given dimensions, you get
  A = 2((12 ft)(6 ft) +(3 ft)(12 ft +6 ft))
  A = 2(72 ft² +(3 ft)(18 ft))
  A = 2(72 ft² +54 ft²) = 252 ft²

The least amount of paper required to cover the box is 252 ft².
4 0
3 years ago
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