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fenix001 [56]
2 years ago
13

Two cars leave towns 760 kilometers apart at the same time and travel toward each other. One car's rate is 12 kilometers per hou

r less than the other's. If they meet in 4 hours, what is the rate of the slower car
Mathematics
1 answer:
zaharov [31]2 years ago
3 0

Answer:

89 km/hr

Step-by-step explanation:

Distance = Rate(speed) × Time

D = R × T

T = 4 hours

One car's rate is 12 kilometers per hour less than the other's.

Hence:

First's car's rate = r × 4 = 4r

Second car's distance = (r + 12) × 4 = 4r + 48

4r + 4r + 48 = 760

8r + 48 = 760

8r = 760 - 48

8r = 712

r = 712/8

r = 89 km/hr

The rate of the slower car is 89 km/hr

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3 years ago
A fence is 10 ft wide by 10 ft long. If posts are placed 2 feet apart how many posts are needed?
yuradex [85]

20

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2 years ago
in triangle PQR the measure of angle R=90, QP=73, RQ=55, and PR=48. what is the value of the sine of angle P to the nearest Hund
alexandr402 [8]

Answer:

P ≈ 48.89°(nearest hundredth)

Step-by-step explanation:

The triangle PQR forms a right angle triangle since angle R is 90°. The triangle has an hypotenuse , adjacent and opposite side.

Using the SOHCAHTOA principle one can find the sine ratio of angle P. Let us designate where each side represent.

opposite side(QR) = 55

adjacent side(PR) = 48

hypotenuse(PQ) = 73

sin P = opposite/hypotenuse

sin P = 55/73

P = sin⁻¹ 55/73

P = sin⁻¹ 0.75342465753

P = 48.8879095605

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3 years ago
Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and pro
Neporo4naja [7]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Area each of them [son and daughter] will get.

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Let, ABCD be the rhombus shaped field and each side of the field be x

[ All sides of the rhombus are equal, therefore we will let the each side of the field be x ]

Now,

• Perimeter = 400m

\longrightarrow  \tt AB+BC+CD+AD=400m

\longrightarrow  \tt x + x + x + x=400

\longrightarrow  \tt 4x=400

\longrightarrow  \tt  \: x =  \dfrac{400}{4}

\longrightarrow  \tt x= \red{100m}

\therefore Each side of the field = <u>100m</u><u>.</u>

Now, we have to find the area each [son and daughter] will get.

So, For \triangle ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

\therefore \tt Simi \:  perimeter \:  [S] =  \boxed{ \sf \dfrac{a + b + c}{2} }

\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}

\longrightarrow \tt S =  \cancel{ \dfrac{360}{2}}

\longrightarrow \tt S = 180m

Using <u>herons formula</u><u>,</u>

\star \tt Area  \: of  \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(80)(80)(20) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180 \times 80 \times 80 \times 20 }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{ {3}^{2} \times  {20}^{2}  \times  {80}^{2}  }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  3 \times 20 \times 80

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} = \red{   4800  \: {m}^{2} }

Thus, area of \triangle ABD = <u>4800 m²</u>

As both the triangles have same sides

So,

Area of \triangle BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

7 0
1 year ago
Read 2 more answers
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