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4 years ago

HELP ME :) please .....

Mathematics

2 answers:

iren [92.7K]4 years ago

8 0

2a^2b^5 is the answer ....

Fantom [35]4 years ago

7 0

They all have a common factor of 2 a^2 and b^5, so the answer is:
2 a^2 b^5

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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 = 100m.

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 herons formula,

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

where

• s is the simi perimeter = 180m

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

Putting the values,

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

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

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

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

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

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

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

Thus, area of $\triangle$ ABD = 4800 m²

As both the triangles have same sides

So,

Area of $\triangle$ BCD = 4800 m²

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

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

★ Figure in attachment.

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

7 0

2 years ago

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Divide 2/8 by 9/18 . Input your answer as a reduced fraction.\

Katen [24]

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{2}{8}\div\frac{9}{18} }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Simplify \ \frac{2}{8} \ to \ \frac{1}{4}. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1}{4}\div\frac{18}{9} }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Use \ this \ rule: a\div \frac{b}{c}=a\times \frac{c}{b}a \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1}{4}\times\frac{18}{9} }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Use \ this \ rule: \frac{a}{b} \times \frac{c}d{}=\frac{ac}{bd}. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1\times18}{4\times9} \ \to \ \ Multiply }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{ \frac{18}{36}= }}\boxed{\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1}{2} }} \end{gathered}$}} \end{gathered}$}$

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2 years ago

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If G is the midpoint of FH, FG= 14x + 25 and GH = 73- 2x, find FH

Alenkinab [10]

Given that G is the midpoint of FH, then:
FG=GH
hence;
14x+25=73-2x
solving for x we get:
14x+2x=73-25
16x=48
x=48/16
x=3
therefore the length FG=14x+25 will be:
14*3+25
=67
hence FH will be:
FH=67*2=134

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3 years ago

Adidea Corp regularly buts merchandise from vendors. It just purchased 1,000 units on credit from one of its vendors. How will t

Lostsunrise [7]

Answer:

You did not include the price they bought the units at.

I will assume this price is $5.

= 5 * 1,000

= $5,000

Recording it will be:

Account Debit Credit

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Accounts Payable $5,000

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3 years ago

Enter the correct answer in the box. Linear function fis shown in the graph. Write the slope-Intercept form of the equation repr

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Answer:

2x+(-7)

Step-by-step explanation:

the point 2 goes through the intercept and -7 goes through the y- intercept

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3 years ago