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Reptile [31]
3 years ago
5

Sum up 40 and 18,subtract therefrom 7.Then multiply with 8.

Mathematics
2 answers:
victus00 [196]3 years ago
8 0
40+18-7×8=58-56=2

or 

(40+18)-7×8=(58)-(56)=2

What exactly do you need help with?
Kryger [21]3 years ago
5 0
[(40+18)-7]\cdot8=(58-7)\cdot8=51\cdot8=408
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Find the value of x of this question ​
mel-nik [20]
<h3><u>Let's</u><u> </u><u>understand</u><u> </u><u>the concept</u><u>:</u><u>-</u></h3>
  • Here angle B is 90°
  • So \triangle ABC and \triangle ABD Are right angled triangle
  • So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
  • First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

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According to Pythagoras thereon

{\boxed{\sf b^2=h^2-p^2}}

  • Substitutethe values

\longrightarrow\sf b^2=10^2-p^2

\longrightarrow\sf b={\sqrt {10^2-8^2}}

\longrightarrow\sf b={\sqrt{100-64}}

\longrightarrow\bf b={\sqrt {36}}

\longrightarrow\sf b=6

\therefore\overline{BC}=6cm

BD=BC+CD

\longrightarrowBD=9+6

\longrightarrowBD=15cm

  • Now in \triangle ABD

Perpendicular=p=8cm

Base =b=15cm

  • We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

{\boxed {\sf h^2=p^2+b^2}}

  • Substitute the values

\longrightarrow\sf h^2=8^2+15^2

\longrightarrow\sf h={\sqrt {8^2+15^2}}

\longrightarrow\sf h={\sqrt {64+225}}

\longrightarrow\sf h={\sqrt {289}}

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\therefore{\underline{\boxed{\bf x=17cm}}}

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3 years ago
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3 years ago
The depth y (in inches) of a lake after x years is represented by the equation y=0.2x+42 . How much does the depth of the lake i
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Answer:

<h2><em><u> 42.8 in</u></em></h2>

Step-by-step explanation:

Step one:

given data

we are told that the expression for the depth of a lake after x years is given by

y=0.2x+42

Required

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Step two:

substitute the value of x in the expression to get y

y=0.2(4)+42\\\\y=0.8+42\\\\y=42.8\\\\

<em><u>After 4 years the depth will increase by 42.8 in</u></em>

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