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2 years ago

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Jane has twice as many books as Kate. if jane gives 14 books to Kate , they will have the same number of books. how many books d

o they have altogether?
need help plss

2 answers:

Sunny_sXe [5.5K]2 years ago

5 0

Hey hope you have a good day today and tomorrow

IRINA_888 [86]2 years ago

4 0

Answer:

I believe the answer is 42 books.

Step-by-step explanation:

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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.

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

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2 years ago

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How to find the area of each of the shaded sectors and round to the nearest hundredth

Mazyrski [523]

Answer:

1-6 you need to write the JHI DEF where the radius center point is in the middle of each shape next to 1-6 some have 2 shapes so use 6 letters there.

Step-by-step explanation:

1. Area = 72 of 360 = 20% of 716.3 = 143.26inch*2 =JHI

2.Area = 19 of 360 = 68.4% of 69.4 = 47.46km^2 =DEF

3. Area = 92 of 360 = 25.56% of 153.9 = 39.33cm^2

4. Area = 226 of 360 = 64.57% of 706.9 = 456.45ft^2

5. Area = 74 of 360 = 20.55% of 153.9 =31.63 inch^2

6. Area = 326 of 360 = 90.56% of 1452 -137.07 = 1314.93m^2

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Your answer is 23.01 in ounces

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3 years ago

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il63 [147K]

Answer:

i think it’s b

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3 years ago

How many planes contain each line and point?

nikitadnepr [17]

There are infinite many planes that contain each line and point

<h3>How to determine the number of planes?</h3>

The given parameters are given as:

Line KL and G
Line JI and G

As a general rule, a line and a point can be used to draw as many planes as possible

This means that there are infinite many planes possible

Hence, there are infinite many planes that contain each line and point

Read more about planes and points at:

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2 years ago