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rosijanka [135]
3 years ago
7

the figure shows secants GH and GK intersecting to form an angle.Find m<1.if necessary ,round to the tenth place.

Mathematics
2 answers:
neonofarm [45]3 years ago
4 0

Answer:

m\angle 1=27^{\circ}

Step-by-step explanation:

We have been given a circle in which secants GH and GK intersecting to form an angle. We are asked to find the measure of angle 1.

We know that the measure of angle formed by two intersecting secants outside the circle is half the difference of intercepted arcs.

m\angle 1=\frac{\widehat{KH}-\widehat{JI}}{2}

m\angle 1=\frac{83^{\circ}-29^{\circ}}{2}

m\angle 1=\frac{54^{\circ}}{2}

m\angle 1=27^{\circ}

Therefore, the measure of angle 1 is 27 degrees.

zysi [14]3 years ago
3 0
M<1 = 1/2(83 - 29)
Therefore, m<1 = 27 degrees.

Hope this helps!
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Determine whether each expression is equivalent to 49^2t – 0.5.
vampirchik [111]

Answer:

None of the expression are equivalent to 49^{(2t - 0.5)}

Step-by-step explanation:

Given

49^{(2t - 0.5)}

Required

Find its equivalents

We start by expanding the given expression

49^{(2t - 0.5)}

Expand 49

(7^2)^{(2t - 0.5)}

7^2^{(2t - 0.5)}

Using laws of indices: (a^m)^n = a^{mn}

7^{(2*2t - 2*0.5)}

7^{(4t - 1)}

This implies that; each of the following options A,B and C must be equivalent to 49^{(2t - 0.5)} or alternatively, 7^{(4t - 1)}

A. \frac{7^{2t}}{49^{0.5}}

Using law of indices which states;

a^{mn} = (a^m)^n

Applying this law to the numerator; we have

\frac{(7^{2})^{t}}{49^{0.5}}

Expand expression in bracket

\frac{(7 * 7)^{t}}{49^{0.5}}

\frac{49^{t}}{49^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{49^{t}}{49^{0.5}} becomes

49^{t-0.5}}

This is not equivalent to 49^{(2t - 0.5)}

B. \frac{49^{2t}}{7^{0.5}}

Expand numerator

\frac{(7*7)^{2t}}{7^{0.5}}

\frac{(7^2)^{2t}}{7^{0.5}}

Using law of indices which states;

(a^m)^n = a^{mn}

Applying this law to the numerator; we have

\frac{7^{2*2t}}{7^{0.5}}

\frac{7^{4t}}{7^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{7^{4t}}{7^{0.5}} = 7^{4t - 0.5}

This is also not equivalent to 49^{(2t - 0.5)}

C. 7^{2t}\ *\ 49^{0.5}

7^{2t}\ *\ (7^2)^{0.5}

7^{2t}\ *\ 7^{2*0.5}

7^{2t}\ *\ 7^{1}

Using law of indices which states;

a^m*a^n = a^{m+n}

7^{2t+ 1}

This is also not equivalent to 49^{(2t - 0.5)}

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3 years ago
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irinina [24]

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