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Nataly_w [17]
3 years ago
7

Can I have halp with all and an explanation because I am terrible at properties.

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
4 0
1.   4 • (–3) • 5
   so 4 x (-3) = -12  and then -12 x 5 = -60
2.   (2.25 x 23) x 4
so (2.25 x 23) 51.75  and then 51.75 x4 = 207
4.  5 x 12 x (-2)
 so 5 x 12 = 60  and then 60 x (-2) = -120
5.   35(26)(0) =
  so 35 x 26 = 910  and then  910 x 0 = 0

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Solve the triangle A = 2 B = 9 C =8
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Answer:

\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}

Then, given the sides a=2, b=9, and c=8.

\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}

For B:

\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}

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Step-by-step explanation:

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PolarNik [594]
4.4x-4=3.4x

Substract '-3.4x' at LHS nad the RHS of the above expression.

\begin{gathered} 4.4x-4-3.4x=3.4x-3.4x \\ 4.4x-3.4x-4=0 \\ x-4=0 \end{gathered}

Add '4' on both LHS and RHs of the above expression.

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4 0
1 year ago
A rectangular building for a gym is three times as long as it is wide. Just inside the walls of the building, there is a 6ft rec
lisov135 [29]

The equation 3x² - 48x + 6856 represents the area of the gym and track together in terms of width.

<h3>What is the area of the rectangle?</h3>

It is defined as the area occupied by the rectangle in two-dimensional planner geometry.

The area of a rectangle can be calculated using the following formula:

Rectangle area = length x width

We have:

A rectangular building for a gym is three times as long as it is wide. Just inside the walls of the building, there is a 6ft rectangular track along the walls of the gym and has an area of 7000ft²

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As the area of track along the walls of the gym is 7000ft²

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After simplifying:

3x² - 48x + 6856 = 0

Thus, the equation 3x² - 48x + 6856 represents the area of the gym and track together in terms of width.

Learn more about the rectangle here:

brainly.com/question/15019502

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