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3 years ago

Find the 76th term of the arithmetic sequence 16, 14, 12, ... I need help pls

Mathematics

2 answers:

Alona [7]3 years ago

6 0

Answer:

<h2>your answer is like this</h2>

Step-by-step explanation:

disa [49]3 years ago

5 0

Answer: The answer is -134

Step-by-step explanation: Common Difference:

d=a_2-a_1=(14)-(16)=-2d=a2−a1=(14)−(16)=−2

a_n=a_1+(n-1)dan=a1+(n−1)d

\text{76th term:}76th term:

a_{76}=16 + (76-1)(-2)a76=16+(76−1)(−2)

a_{76}=16 + (75)(-2)a76=16+(75)(−2)

a_{76}=-134a76=−134

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yanalaym [24]

Answer:

Step-by-step explanation:

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4 years ago

Answer this problem 3/14 divided 2/7<br><br> a. 2/7<br><br> b. 1/2<br><br> c. 1/7<br><br> d. 3/4

Ierofanga [76]

Answer:

Step-by-step explanation:

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3 years ago

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8 0

3 years ago

Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest to find Maya Rodriguez.

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Answer:

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2 years ago

Rationalize the denominator.

Maslowich

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166412

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$\mathsf{\dfrac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{\sqrt{a}+2\sqrt{y}:}$

$\mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}\\\\\\ \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)^2}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}$

The numerator is a square of a sum, and the denominator is the product of a difference and a sum (product of two conjugates). Then, expand the common products:

$\mathsf{=\dfrac{\big(\sqrt{a}\big)^2+2\cdot \sqrt{a}\cdot 2\sqrt{y}+\big(2\sqrt{y}\big)^2}{\big(\sqrt{a}\big)^2-\big(2\sqrt{y}\big)^2}}\\\\\\ \mathsf{=\dfrac{a+4\cdot \sqrt{a}\cdot \sqrt{y}+2^2\cdot \big(\sqrt{y}\big)^2}{a-2^2\cdot \big(\sqrt{y}\big)^2}}$

$\mathsf{=\dfrac{a+4\sqrt{ay}+4y}{a-4y}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

8 0

3 years ago