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Levart [38]
2 years ago
11

In circle K with the measure of minor arc \stackrel{\Large \frown}{JL}=48^{\circ}, JL ⌢ =48 ∘ , find \text{m} \angle JML.m∠JML.

Mathematics
1 answer:
Stells [14]2 years ago
3 0

Answer:

2.5

Step-by-step explanation:

from delta math

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Nat2105 [25]

Answer:

x

Step-by-step explanation:

1 - \frac{x}{2} > -4

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7 0
2 years ago
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Evaluate the expression for n = –8.<br><br> –2n − 6 =
Nana76 [90]

Answer:

<h2>10</h2>

Step-by-step explanation:

<h3>-2n - 6 = ?</h3><h3 /><h3>let n be -8</h3><h3 /><h3>-2 (-8) - 6 = ?</h3><h3 /><h3>= 10</h3>

\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}

5 0
3 years ago
I need help plez and thanks
Elenna [48]
The answer is 80, there multiplying x2 for each number
7 0
3 years ago
I need help if you can that would be great thank you
liq [111]

Answer:

Simplifying the expression (3\:.\:2)^5\div (3^2\:.\:2^3) so, there is only one power of each base we get \mathbf{3^3\:.\:2^2}

Option D is correct answer.

Step-by-step explanation:

We need to simply the expression (3\:.\:2)^5\div (3^2\:.\:2^3) so, there is only one power of each base.

Solving:

(3\:.\:2)^5\div (3^2\:.\:2^3)

We can write it as:

\frac{(3\:.\:2)^5}{3^2.2^3}

Now using exponent rule: (a\:.\:b)^m=a^m\:.\:b^n

\frac{3^5\:.\:2^5}{3^2.2^3}

Now using the exponent rule: \frac{a^m}{a^n}=a^{m-n} the bases should be same

3^{5-2}\:.\:2^{5-3}\\=3^3\:.\:2^2

So, simplifying the expression (3\:.\:2)^5\div (3^2\:.\:2^3) so, there is only one power of each base we get \mathbf{3^3\:.\:2^2}

Option D is correct answer.

5 0
3 years ago
Draw a related picture for each of the following<br>​
exis [7]

Answer:

what are the following ?

Step-by-step explanation:

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2 years ago
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