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kirill [66]
3 years ago
15

The regular selling price of an item is $261. For special year-end sale the price is at a markdown of 20%. Find the discount pri

ce.
Mathematics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

208.8 or 209 for the discount price

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8 trees increased by 175%
fredd [130]

Answer:

14

Step-by-step explanation:

8x175%= 14

6 0
3 years ago
Homework Week 13 and 14<br> Introduction to Inequalities
irina [24]

Answer:

you are going to have to look from left to right and see if it is above or below

7 0
3 years ago
Write the expanded form of the decimal 0.082¯ using powers of 10
Softa [21]

Answer:

0.082 = 8.2× 10^-2

Step-by-step explanation:

Given the decimal number 0.082, the expanded form using the power of 10 can be gotten by writing the decimal number in standard format(writing as a multiple of 10).

To do that we will shift the decimal point to the front up to the front of digit 8. This shows that the decimal point will be shifted 2times to the front. Since of is shifted 2times to the front, our power of 10 will be -2.

0.082 = 8.2× 10^-2

This gives the required answer.

Note that, the power of 10 is positive when decimals are shifted to the back and negative when shifted to the front(in this case).

3 0
3 years ago
The 13th term of a geometric sequence is 16, 384 and the first term is 4. What is the common ratio?
AleksAgata [21]

Answer:

\large \boxed{2}

Step-by-step explanation:

The formula for the nth term of a geometric sequence is

aₙ = a₁rⁿ⁻¹

In your geometric sequence, a₁ = 4 and a₁₃ = 16 384.

\begin{array}{rcl}16384 & = & 4r^{(13 - 1)}}\\16384 & = & 4r^{12}\\4096 & = & r^{12}\\3.6124 & = & 12 \log r\\0.30102 & = & \log r\\r & = & 10^{0.30102}\\ & = & \mathbf{2}\\\end{array}\\\text{The common ratio is $\large \boxed{\mathbf{2}}$}

Check:

\begin{array}{rcl}16384 & = & 4(2)^{12}\\16384 & = & 4(4096)\\16384 & = & 16384\\\end{array}

It checks.

7 0
3 years ago
For which values of p and q, will the following pair of linear equations have infinitely many
Tema [17]
In order to have infinitely many solutions with linear equations/functions, the two equations have to be the same;
In accordance, we can say:
(2p + 7q)x = 4x [1]
(p + 8q)y = 5y [2]
2q - p + 1 = 2 [3]
All we have to do is choose two equations and solve them simultaneously (The simplest ones for what I'm doing and hence the ones I'm going to use are [3] and [2]):
Rearrange in terms of p:
p + 8q = 5 [2]
p = 5 - 8q [2]
p + 2 = 2q + 1 [3]
p = 2q - 1 [3]
Now equate rearranged [2] and [3] and solve for q:
5 - 8q = 2q - 1
10q = 6
q = 6/10 = 3/5 = 0.6

Now, substitute q-value into rearranges equations [2] or [3] to get p:
p = 2(3/5) - 1
p = 6/5 - 1
p = 1/5 = 0.2
3 0
3 years ago
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