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3 years ago

Given: m∠1 = 95m∠2 = 36∠3 ≅ ∠ACE please help

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

3 0

The measure of the unknown angles are as listed:

m∠3 = 85degrees
m∠4 = 85degrees
m∠6 = 36degrees
m∠5 = 144degrees
m∠7 = 59degrees

Given the following parameters:

m∠1 = 95 degrees

m∠2 = 36 degrees

Since the sum of angle on a straight line is 180 degrees, hence;

m∠1 +m∠3 = 180

95 + m∠3 = 180

m∠3 = 180 - 95

m∠3 = 85degrees

From the diagram, m∠3 = m∠4 = 85 degrees (vertical angles)

Similarly, m∠6 = m∠2 = 36degrees (vertical angles)

The sum of angle on a straight line is 180 degrees, hence;

m∠2 +m∠5 = 180

36 + m∠5= 180

m∠5 = 180 - 36

m∠5 = 144degrees

Taking the sum of the interior angle of the triangle BEF;

m∠4 + m∠6 + m∠7 = 180

85 + 36 + m∠7 = 180

121 + m∠7 = 180

m∠7 = 180 - 121

m∠7 = 59degrees

Learn more on angles here: brainly.com/question/25716982

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The digit 5 appears twice in the number 255,120. How does the total value of the 5 on the right compare to the total value of th

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Answer:

The value of the first " $5$ " in the number $255,\!120$ is ten times that of the second " $5\!$ " in this number.

Step-by-step explanation:

What gives the number " $255,\!120$ " its value? Of course, each of its six digits has contributed. However, their significance are not exactly the same. For example, changing the first $\verb!5!$ to $\verb!6!$ would give $2\mathbf{6}5,\!120$ and increase the value of this number by $10,\!000$ . On the other hand, changing the second $\verb!5!\!$ to $\verb!6!\!$ would give $25\mathbf{6},\!120$ , which is an increase of only $1,\!000$ compared to the original number.

The order of these two digits matter because the number " $255,\!120$ " is written using positional notation. In this notation, the position of each digits gives the digit a unique weight. For example, in $255,\!120\!$ :

$\begin{array}{|r||c|c|c|c|c|c|}\cline{1-7}\verb!Digit!& \verb!2! & \verb!5! & \verb!5! & \verb!1! & \verb!2! & \verb!0!\\\cline{1-7}\textsf{Index} & 5 & 4 & 3 & 2 & 1& 0 \\ \cline{1-7} \textsf{Weight} & 10^{5} & 10^{4} & 10^{3} & 10^{2} & 10^{1} & 10^{0}\\\cline{1-7}\end{array}$ .

(Note that the index starts at $0$ from the right-hand side.)

Using these weights, the value $255,\!120$ can be written as the sum:

$\begin{aligned}& 255,\!120\\ &= 2 \times 10^{5} + 5 \times 10^{4} + 5 \times 10^{3} + 1 \times 10^{2} + 2 \times 10^{1} + 0 \times 10^{0} \\&=200,\!000 + 50,\!000 + 5,\!000 + 100 + 20 + 0 \end{aligned}$ .