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Darina [25.2K]
3 years ago
7

Determine the area 9 cm 13 cm 12 cm 14cm

Mathematics
1 answer:
Maslowich3 years ago
5 0
140.5 square cm is the answer.
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Can anyone help with this question
4vir4ik [10]

Answer: A) x=6

Step-by-step explanation: my trick that i do with these questions are to just fill in the x's with the different choices that they give you.

so... first try out 6. fill in every x with a 6. so now, x isn't x anymore, it's 6.

(8-6)=2

9 times 2= 18

3 times 6= 18

this means that x would equal 6

if that first one doesn't work, try out the other ones. but this one works, so we know we have the right answer. i hope this helps with this problem, and i hope it helps with other problems like this too.

:)

3 0
4 years ago
WILL MARK BRAINLIEST!! 5 POINTS!!
Anton [14]
The speed of the car per minute is 3,520 feet (choice c). This is because in order to solve the problem, you have to convert 40 mi/h into feet. Since there are 5,280 feet per mile, you have to do 5,280 x 40, which is 211,200, which is the number of feet per hour. But, the question is asking for ft/min. There are 60 minutes in an hour, so you divide 211,200 by 60, which is 3,520. So, the speed of the car per minute is 3,520 feet.
5 0
3 years ago
Read 2 more answers
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Yuliya22 [10]

Answer:

C

Step-by-step explanation:

There are 6 units in the numberline, and 2 are shaded. So, 2/6 are shaded.

8 0
3 years ago
Sales of 10000 increased by 65% each year write a function that represents the situation
erastovalidia [21]

Answer: y=10000(1.65)^x


Step-by-step explanation:

Given: The sales of 10000 increased by 65% each year .

Let x be the number of years.

The rate of growth=65%= 0.65

The exponential growth function is given by

y=A(1+r)^x

, where A is initial amount and r is the rate of growth in decimal.

Then the increase in the sales id given by :-

y=10000(1+0.65)^x\\\Rightarrow\ y=10000(1.65)^x

6 0
3 years ago
I have the first part but I don't know how to do the second part<br><br> h e l p
Debora [2.8K]

\bf \begin{array}{cll} \stackrel{ordinal}{position}&\stackrel{term}{value}\\ \cline{1-2} 1&12\\ 2&12+d\\ 3&12+d+d\\ 4&12+d+d+d\\ &12+3d \end{array}\qquad \implies ~\hfill \begin{array}{llll} \stackrel{\textit{4th term}}{12+3d}~~=~~\stackrel{\textit{4th value}}{3}\\\\ 3d=-9\implies d = \cfrac{-9}{3}\\\\ \stackrel{\textit{common difference}}{d = -3} \end{array}

well, we know the common difference is -3, to go from the 4th term to the 8th term, we need to add "d" 4 times or namely 3+4(-3), likewise to go from the 13th term to the 19th term we have to add "d" 6 times, or namely -24 + 6(-3).

\bf \stackrel{\textit{8th term}}{3+4(-3)}\implies 3-12\implies -9 \\\\\\ \stackrel{\textit{19th term}}{-24 + 6(-3)}\implies -24-18\implies -42

3 0
3 years ago
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