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2 years ago

Free brainlest if u help pls

Mathematics

1 answer:

Oliga [24]2 years ago

5 0

5x/4-2=3x/2-4 is true

5x/4 -2=6x/4-4

5x/4+2=6x/4

-5x/4

2=x/4

x=8

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A scale drawing of a square has a side length of 9 inches. The drawing has a scale of 1 in. : 5 mi. Find the actual perimeter an

Dennis_Churaev [7]

Answer:

Area = 2025 miles²

Perimeter = 180 miles

Step-by-step explanation:

Given the following :

Scale of drawing = 1 inch : 5 miles

Length of side of the scale drawing = 9 inches

To find The actual perimeter and area of the square, we need ;

Actual length of the drawing :

1 inch = 5 miles

9 inches = (5 × 9) miles = 45 miles

Hence,

Area of a square = a²

Where a = side length

Actual side length = 45 miles

Actual area of square = 45² = 2025 miles²

Actual Perimeter of square :

Perimeter of a square = 4a

a = side length

Actual perimeter = 4(45) = 180 miles

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Answer:

Step-by-step explanation:

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2 years ago

Nathan wants to draw a triangle. he knows that two of the side lengths are 5 inches and 7inches. What is the possible length for

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2 years ago

State whether the given measurements determine zero, one, or two triangles. b=86, b=21, c=18

hammer [34]

Answer: 1

Step-by-step explanation:

I assume you meant to type $B=86^{\circ}, b=21, c=18$ .

By the Law of Sines,

$\frac{\sin C}{c}=\frac{\sin B}{b} \\ \\ \frac{\sin C}{18}=\frac{\sin 86^{\circ}}{21} \\ \\ \sin C=\frac{18 \sin 86^{\circ}}{21} \\ \\ C=\sin^{-1} \left(\frac{18 \sin 86^{\circ}}{21} \right), 180^{\circ}-\sin^{-1} \left(\frac{18 \sin 86^{\circ}}{21} \right) \\ \\ C \approx 58.77^{\circ}, 121.23^{\circ}$

Since only one of these values will make a triangle (the obtuse possibility for C will mean $B+C > 180^{\circ}$ , which violates the condition that the angles of a triangle add to 180 degrees), 1 triangle can be formed.