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3 years ago

12

A(1) = -13 a(n-1) + 4 find the 2nd term in the sequence

1 answer:

kirill115 [55]3 years ago

5 0

Solution

The sequence is an Arithmetic sequence.

Here the 1st term = -13

An = A(n-1) + 4

Therefore, the common difference (d) = 4

2nd term = first term + common difference

2nd term = -13 + 4 = -9

The answer is -9.

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For what values of x is the following equation satisfied : 3x+9 = 21+7x ?

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First of all, the first power of 'x' is the highest power in the whole equation,
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You said that <span><u>3x+9 = 21+7x </u>

Subtract 3x from each side:

9 = 21 + 4x

Subtract 21 from each side:

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Answer for brainliest

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Answer:

red area = 7.74 in^2

Step-by-step explanation:

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At Shop A, 2 pounds of crawfish costs $6. At Shop B, 5 pounds of crawfish costs $15. What is the unit price for crawfish at Shop

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3 years ago

The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?

Marrrta [24]

Answer:

Is an acute triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

G(7,3), H(9, 0), I(5, -1)

step 1

Find the distance GH

substitute in the formula

$d=\sqrt{(0-3)^{2}+(9-7)^{2}}$

$d=\sqrt{(-3)^{2}+(2)^{2}}$

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step 2

Find the distance IH

substitute in the formula

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$d=\sqrt{(1)^{2}+(4)^{2}}$

$IH=\sqrt{17}\ units$

step 3

Find the distance GI

substitute in the formula

$d=\sqrt{(-1-3)^{2}+(5-7)^{2}}$

$d=\sqrt{(-4)^{2}+(-2)^{2}}$

$GI=\sqrt{20}\ units$

step 4

Verify what type of triangle is the polygon

we know that

If applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$ ----> is a right triangle

$c^{2}> a^{2}+b^{2}$ ----> is an obtuse triangle

$c^{2}< a^{2}+b^{2}$ ----> is an acute triangle

where

c is the greater side

we have

$c=\sqrt{20}\ units$

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substitute

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