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3 years ago

The first term of an arithmetic sequence is –12 and the ninth

Mathematics

1 answer:

shtirl [24]3 years ago

8 0

Answer:

d = $\frac{7}{2}$

Step-by-step explanation:

The nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₉ = 16 , then

- 12 + 8d = 16 ( add 12 to both sides )

8d = 28 ( divide both sides by 8 )

d = $\frac{28}{8}$ = $\frac{7}{2}$

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20 points! Will mark the brainliest!

andrew-mc [135]

$\dfrac{ \dfrac{x+3}{4x^2-16} }{ \dfrac{2x^2+10x+12}{2x-4} }$

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Write the divide fraction horizontally:
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$= \dfrac{x+3}{4x^2-16} \div \dfrac{2x^2+10x+12}{2x-4}$

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Factorise the numerators and denominators when possible:
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$= \dfrac{x+3}{4(x+2)(x-2)} \div \dfrac{ 2(x + 3) (x + 2)}{2(x-2)}$

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Convert the divide fraction to multiplication fraction
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$= \dfrac{x+3}{4(x+2)(x-2)} \times \dfrac{2(x-2)}{2(x+3)(x+2)}$

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Cancel the factors
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$= \dfrac{1}{4(x+2)} \times \dfrac{1}{(x+2)}$

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Combine to single fraction
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$= \dfrac{1}{4(x+2)^2}$

4 0

4 years ago

Add. reduce to lowest terms. 6 4/5 + 3 7/10=

KiRa [710]

I would like to help but i dont know either

7 0

3 years ago

While driving from Chicago, Brent had a visibility problem due to the fog. After 1 1/2

liq [111]

1) Let x be the original speed that he was driving;

x mph ---> 1.5 hours
x+44 mph --> 2.5 hours

1.5x + 2.5(x+44)=206
1.5x+2.5x+110=206
4x=206-110
4x=96
x=24
Therefore, he was travelling at 24 mph in the fog.

2) Let x be the speed of 1 snail,
The other snail travels at the speed of x+1 cm/min

23x+23(x+1)=391 <-- solve for x
23x+23x+23=391
46x=368
x=368/46
x=8

Therefore one of the snail travels at 8 cm/min while the other goes at 9 cm/min

7) Let x be one crew
Then the other crew would be working at x+0.5

3.2x+3.2(x0.5)=8 <-- solve for x
3.2x+1.6x=8
4.8x=8
x=8/4.8
x=1.6666 or 1.7 miles

Therefore, one of the crew does 1.7 miles per day and the other crew does (0.5*1.7)=0.85 miles

Hope I helped :)

5 0

4 years ago

the longest side of an acute isosceles triangle 8 centimeters . Round to the nearest tenth, what is the smallest possible length

Triss [41]

Let x be the length of each of two congruent sides.

The triangle will be аcute if:
x² + x² > 8²
2x² > 64
x² > 32
x > √32
x > 5.657

So, the smallest possible length of one of two congruent sides have to be 5.7 cm (to the nearest tenth)

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3 years ago

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