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3 years ago

The sum of 18 and twice a number use n to rep Repesent unknown number

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

7 0

18+2n. For sure hope it helps

Ivanshal [37]3 years ago

7 0

18+2n. Hope this helps!

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Ray is buying some ginger roots to brew some fresh ginger ale. The price of the ginger root is g, and Ray has a coupon for 10% o

kondaur [170]

Step-by-step explanation:

Comparision: Let say you buy chocolate that cost $1. And there was luckily a coupon, for 10% off.

That means after applying the coupon, you will only pay 90% of the original price. Which is

$0.9(1) = 0.9$

which equals 90 cents.

So let go back to this problem and apply similar thinking.

The price of the original root is g, so for the original price column, check the 3rd cell, g.

Use our comparision problem, by applying the coupon to the chocolate, you will save

$1 - 0.9 = 0.1$

which is 1/10 of the original price.

In this equation, we can represent this as 0.1g.

So for the second column, shade in 5th cell, 0.1g.

The price would be

$(0.9)1 = 0.9$

So our equation for this equation will be

0.9g.

So for third column, shade in 1st cell, 0.9g.

7 0

2 years ago

Which set of data has the weakest linear association?

motikmotik

The top left-hand corner set of data has the weakest linear association.

Hope that helped!!! :)

5 0

3 years ago

PLEASE HELP

k0ka [10]

Answer:

1 = 0.2

2 = 32

3 = 0.33...

4 = 64

5 = 10,000

6 = 7

8 0

2 years ago

A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequ

andreyandreev [35.5K]

The correct answer is C) t₁ = 375, $t_n=5t_{n-1}$ .

From the general form,
$t_n=75(5)^n$ , we must work backward to find t₁.

The general form is derived from the explicit form, which is
$t_n=t_1(r)^{n-1}$ . We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

$t_n=t_1(5)^{n-1}$

Using the products of exponents, we can "split up" the exponent:
$t_n=t_1(5)^n(5)^{-1}$

We know that 5⁻¹ = 1/5, so this gives us
$t_n=t_1(\frac{1}{5})(5)^n
\\
\\=\frac{t_1}{5}(5)^n$

Comparing this to our general form, we see that
$\frac{t_1}{5}=75$

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
$t_n=t_{n-1}(r)$ , while we must state what t₁ is; this gives us

$t_1=375; t_n=t_{n-1}(5)$

3 0

3 years ago

Please help with calculations!

ASHA 777 [7]

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