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Marina86 [1]
3 years ago
5

Which student's work is correct? Explain please

Mathematics
1 answer:
Lunna [17]3 years ago
8 0

Step-by-step explanation:

Wendy

-20 = x/2 + 7

-20 -7 = x/2 + 7 -7

-27 = x/2

-54 = x

x = -54

because it's math rule

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The answer should be c because the independent variable should not be manipluated. 
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What is the solution of (8x-8)3/2=64
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Well x=19/3
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Read 2 more answers
A 4 meter long ladder leans against a 90 degree wall at a 68 degree angle.
Alenkinab [10]

Answer:

Step-by-step explanation:

We use trig to answer these questions:

sin x = opposite side of a triangle / hypotenuse

cos x = adjacent side / hypotenuse

Construct a right triangle with hypotenuse 4 m and lower angle 68 degrees.

The height of the top of the ladder  is then defined by:

sin 68 degrees = height / 4 m, which yields (4 m)(sin 68) = height =

3.71 m

and the distance of the bottom of the ladder from the wall as follows:

cos 68 degrees = horizontal distance / hypotenuse =

1.50 m

3 0
2 years ago
Based on the diagram shown, find θ to the nearest degree.
sattari [20]

Answer:

  θ = 38°

Step-by-step explanation:

The lower right triangle is congruent to the upper left triangle, so we have θ and 20° being the two acute angles in the triangle. The law of sines tells you ...

  sin(θ)/9 = sin(20°)/5

  sin(θ) = (9/5)sin(20°)

  θ = arcsin(9/5·sin(20°)) ≈ 38°

___

Another solution to the triangle is θ = 180° -38° = 142°. The diagram clearly shows θ as an acute angle, so we take this second solution to be extraneous.

8 0
3 years ago
Explain how to multiply the following whole numbers 21 x 14
Lesechka [4]

Answer:

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Step-by-step explanation:

Given

21\:\times \:14

Line up the numbers

\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

Multiply the bold numbers:    1×4=4

\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}

Multiply the bold numbers:    2×4=8

\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the top number by the bolded digit of the bottom number

\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the bold numbers:    1×1=1

\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}

Multiply the bold numbers:    2×1=2

\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}

Add the rows to get the answer. For simplicity, fill in trailing zeros.

\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}

adding portion

\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}

Add the digits of the right-most column: 4+0=4

\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}

Add the digits of the right-most column: 8+1=9

\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}

Add the digits of the right-most column: 0+2=2

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Therefore,

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

6 0
3 years ago
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