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erastova [34]
2 years ago
13

Brainlist! in what ways could the shape pattern below be described?​

Mathematics
1 answer:
creativ13 [48]2 years ago
7 0
A . the 13th spot will have a star
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Evaluate the expression 50 + 0.5 *(41-32)
kow [346]
PEMDAS
P- 41-32 = 9
M- 0.5 * 9 = 4.5
A - 50 +4.5 = 54.5
54.5 is your answer
6 0
3 years ago
Complete the table below?
siniylev [52]

ABC = $425

Load = 0

Total Cost = $425

Sales Price = $850

Sales price ÷ total cost = 850/425

= 2%

DEF = $600

Load = $375

Total Cost = $975

Sales Price = $1200

Sales Price ÷ Total cost = 1200/975

= 1% (Nearest 1%)

Must click thanks and mark brainliest

5 0
3 years ago
(12) In a relay race, Mia ran 1.43 kilometers. Then she passed the baton to Gerald, who ran an
BigorU [14]

Answer:

4.13

Step-by-step explanation:

To find out how many kilometres they ran in all you have to do 1.43+2.7 =4.13

3 0
3 years ago
Read 2 more answers
Find the amount in an account if 7,650 is invested at 9.15 percent compounded quarterly for 8 years and 6 months
storchak [24]

We have been given that in an account an amount of 7,650 is invested at 9.15 percent compounded quarterly for 8 years and 6 months.

We will use compound interest formula to find our answer.

A=Pcdot(1+\frac{r}{n}) ^{nT},

Where, P= principle amount, A= amount after T years, n= period of compounding and r = interest rate (decimal).

Let us substitute our given values in our formula.

A=7650(1+\frac{0.0915}{4} )^{(8.5\times4)}  

A=7650(1+0.022875)^{34}

A=7650(1.022875)^{34}  

A=7650\cdot 2.15758136398

A=16505.4974    

Therefore, after 8 years and 6 months our amount will be 16505.497.


7 0
3 years ago
Plzz anyone solve all answers plzzzzzzzz​
algol13

You posted a lot of problems here. In the future please only post one problem at a time. Thank you.

I'll do the first two problems to get you started. Hopefully it will help you finish off the rest of the questions.

==========================================

Problem 1

{18, a, b, -3} is an arithmetic sequence or arithmetic progression (AP).

This means we have some number d added on to each term to get the next term.

first term = 18

second term = first term + d = 18+d = a

third term = second term + d = (18+d)+d = 18+2d = b

fourth term = third term + d = (18+2d)+d = 18+3d = -3

----

Let's solve that last equation for d

18+3d = -3

18+3d-18 = -3-18

3d = -21

3d/3 = -21/3

d = -7

----

The value d = -7 tells us to add -7 to each term to get the next term. In other words, we subtract 7 from each term to get the next term

first term = 18

second term = first term + d = 18+d = 18+(-7) = 18-7 = 11

third term = second term + d = 11+d = 11+(-7) = 11-7 = 4

fourth term = third term + d = 4+d = 4+(-7) = 4-7 = -3

----

We see that a = 11 and b = 4 are the second and third terms respectively.

Therefore, a+b = 11+4 = 15

-------------

<h3>Answer: 15</h3>

==========================================

Problem 2

A multiple of 4 is in the form 4*n for some integer n, ie n is a whole number.

We want to know which values of 4*n are between 10 and 250.

----

Divide both 10 and 250 by 4 to get the following

10/4 = 2.5

250/4 = 62.5

If n = 2, then 4*n = 4*2 = 8 is not between 10 and 250; however n = 3 will make 4*n = 4*3 = 12 to be between 10 and 250. We see that n = 3 is the smallest possible allowed value.

If n = 62, then 4*n = 4*62 = 248 is between 10 and 250; while n = 63 will make 4*n too big because 4*63 = 252. The largest n can get is n = 62

----

The question posed in question 2 is equivalent to asking the following: "How many values are in the set {3, 4, 5, ..., 60, 61, 62}?"

You could count all of the values in the set, but that exercise is very tedious busywork. There's a much faster way. First lets consider the set below

{a, a+1, a+2, ..., b-2, b-1, b}

where a,b are integers. Basically this set starts at 'a', counts up until we get to 'b'. The handy formula

c = b-a+1

will provide the exact count of values in the set {a, a+1, a+2, ..., b-2, b-1, b}

----

In this case, a = 3 and b = 62, making

c = b-a+1

c = 62-3+1

c = 60

There are 60 values in the set {3, 4, 5, ..., 60, 61, 62}

There are 60 multiples of four that are between 10 and 250.

-------------

<h3>Answer: 60</h3>
4 0
3 years ago
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