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mamaluj [8]
3 years ago
8

Guys please give me the answer for this q it's so hard

Mathematics
2 answers:
posledela3 years ago
5 0

Answer:

\frac{9}{20}

Step-by-step explanation:

change the mixed number to an improper fraction

2 \frac{1}{4} = \frac{9}{4}

We have to calculate

\frac{1}{3} × \frac{3}{5} × \frac{9}{4}

simplify by cancelling the 3's on the numerator/denominator

then

= 1 × \frac{1}{5} × \frac{9}{4}

multiply the remaining values on the numerators /denominators

= \frac{9}{5(4)} = \frac{9}{20}


rusak2 [61]3 years ago
3 0

Answer:

9/20

Step-by-step explanation:

1/3 *  3/5 * 2 1/4 =

1/3 * 3/5 * 9/4=

1/5 * 9/4 =

9/20


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Answer:

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Step-by-step explanation:

In f(x)=a\cos(b(x-c))+d, b represents a constant related to the period of the function. Here's how it's related:

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b=\frac{2\pi}{12\pi}=\boxed{\frac{1}{6}}

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Please help: linear algebra problem. (Linear combinations)
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Answer:

\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

This tells us that:

A=\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right]

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Step-by-step explanation:

So we are saying we have scalars, c and d, such that:

c\left[\begin{array}{ccc}5\\5\\ 3\end{array}\right]+d\left[\begin{array}{ccc}7\\-8\\-9\end{array}\right]=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right].

So we want to find a way to express this as:

Ax=b where x is the scalar vector, \left[\begin{array}{ccc}c\\d\end{array}\right].

So we can write this as:

\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

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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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