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3 years ago

17. What is 43 in expanded form?

Mathematics

2 answers:

adell [148]3 years ago

8 0

Hey there, none of these can be expanded to 43

masha68 [24]3 years ago

3 0

Answer:

None of those that you listed are correct.

Step-by-step explanation:

All of the selections you've shown do not equal to 43. Sorry. :(

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Select the correct answer from each drop-down menu. Ryan's new employer has offered him a few insurance plans as part of his emp

alex41 [277]

Answer:

Ryan's new employer has offered him a few insurance plans as part of his employee benefits package.

Step-by-step explanation:

Ryan is working as deputy manager in a company. He currently has salary and there are no additional benefits. He is currently having only health insurance. When he joins the new company he gets few insurance plans as part of employee benefit packages. This is additional benefits which attracts a candidate to join a company and it keeps the employees motivated.

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3 years ago

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Answer:

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3 years ago

Select the lesser of The two given numbers -152, 12

Ymorist [56]

Answer:

-152

Step-by-step explanation:

-152 < 12

-152 is negative, which means it is less than 1.

12 is positive, which means it is greater than 1.

Therefore, -152 is the lesser of the two given numbers.

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1 year ago

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To find the measure of ∠A in ∆ABC, use the___(Pythagorean Theorem, Law of Sines, Law of Cosines). To find the length of side HI

nadya68 [22]

Part 1) To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

$\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}$

in this problem we have

$\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})$

therefore

the answer Part 1) is

Law of Sines

Part 2) To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

$g^{2}=h^{2}+i^{2}-2*h*i*cos\ G$

In this problem we have

g=HI

G=angle Beta

substitute

$HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta$

$HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}$

therefore

the answer Part 2) is

Law of Cosines

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3 years ago

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Simplify the square root of 3 times the fifth root of 3.

bogdanovich [222]

Answer:

<h2>Three to the three fifths power.</h2>

Step-by-step explanation:

The given expression is

$\sqrt{3\sqrt[5]{3} }$

To simplify this expression, we have to use a specific power property which allow us to transform a root into a power with a fractional exponent, the property states:

$\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$

Applying the property, we have:

$\sqrt{3\sqrt[5]{3}}=\sqrt{3(3)^{\frac{1}{5}}}=(3(3)^{\frac{1}{5}})^{\frac{1}{2}}$

Now, we multiply exponents:

$(3(3)^{\frac{1}{5}})^{\frac{1}{2}}\\3^{\frac{1}{2}}3^{\frac{1}{10}}$

Then, we sum exponents to get the simplest form:

$3^{\frac{1}{2}}3^{\frac{1}{10}}=3^{\frac{1}{2}+\frac{1}{10}} =3^{\frac{10+2}{20}}=3^{\frac{12}{20}} \\\therefore \sqrt{3\sqrt[5]{3}}=3^{\frac{3}{5} }$

Therefore, the right answer is three to the three fifths power.

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3 years ago

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