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2 years ago

7

Write the following phrase as an algebraic expression. Use n for “a number.” the product of two and the sum of a number and eigh

t
a. 2(n -8)
b.n+8/2
c. 2n + 8
d. 2(n + 8)

2 answers:

Lemur [1.5K]2 years ago

8 0

We are to write the expression for:

<span>The product of two and The sum of a number and eight
</span>
This means 2 is to be multiplied with the sum of 8 and the number. The number is n. So sum of 8 and number is 8 + n

Product of 2 and the sum is 2 (8 + n)

Therefore, the correct answer is option D

Y_Kistochka [10]2 years ago

4 0

the correct answer is D

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$\tt{ Hey \: there , \: Mr.Panda \: ! }$ ;)

♨ $\large{ \tt{ E \: X \: P \: L \: A \: N \: A \: T \: I\: O \: N}}:$

⤻ Before solving the given question , you should know the answer of these questions :

✺How do you find the hypotenuse , perpendicular and base when the angle ( $\theta \: , \alpha \: ,\beta$ ) is given ?

⇾ The longest side , which is the opposite side of right angle is the hypotenuse ( h ). There are two other sides , the opposite and the adjacent. The naming of these sides depends upon which angle is involved. The opposite is the side opposite the angle involved and it is called the perpendicular ( p ) . The adjacent us the side next to the angle involved ( buy not the hypotenuse ) and it is called the base ( b ).

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In the above cases , $\theta$ is taken as the angle of reference.

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❈ $\large{ \tt{GIVEN}} :$

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✧ $\large{ \tt{TO \: FIND} : }$

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✎ $\large{ \tt{SOLUTION}} :$

Firstly , Finding the value of perpendicular ( p ) using Pythagoras theorem :

❃ $\boxed{ \sf{ {h}^{2} = {p}^{2} + {b}^{2} }}$ [ Pythagoras theorem ]

$\large{ ⇢ \sf{p}^{2} + {b}^{2} = {h}^{2} }$

$\large{⇢ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}$

$\large{ ⇢\sf{ {p}^{2} = {18}^{2} - {16}^{2} }}$

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$\large{ \tt{ ↬ \boxed{ \tt{tan \theta = \frac{ \sqrt{17} }{8}}}}}$

⟿ $\boxed{ \boxed{ \tt{OUR\: FINAL \: ANSWER : \boxed{ \underline{ \bf{ \frac{ \sqrt{17} }{8}}}}}}}$

۵ Yay! We're done!

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