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Pani-rosa [81]
4 years ago
9

A small bag of flavor weighed 13 ounces. A large bag was 5 percent heavier.How much does the large bag weigh?

Mathematics
1 answer:
Novay_Z [31]4 years ago
5 0
13+5=18
The large bag would be 18 because you have to add 5 more ounces than 13
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At 9:00 pm bob leaves his home and starts walking to the bus stop at a constant speed of 9.3 feet per second. bob's house is 432
klio [65]
a. define variables for the quantities that are changing (be specific).
 We define variables:
 t = time in second
 d = distance traveled
 The variables that are changing are t and d.
 As the time increases, the distance decreases.

 b. what does it mean to say bob walks at a constant speed of 9.3 feet per second?
 It means that for every second that passes, Bob walks 9.3 feet.

 c. consider the formula d = 4320 - 9.3t i. what does t represent? ii. what does 9.3t represent? iii. what does d represent? 
 For this case we have:
 t = represents the time in seconds
 9.3t = represents the distance traveled by Bob
 d = represents the distance Bob must travel to get to the bus stop.
5 0
3 years ago
Find the values of a and b so that the solution of the linear system is (-9,1).
Diano4ka-milaya [45]

Answer:

Step-by-step explanation:

we can write -9 instead of x and y=1 instead of 1

so we write solution again

-9a+1b=-31

-9a-1b=-41

-18a=-72

a=4

we should write 4 instead of a

-9(4)+1b=-31

-36+b=-31

b=5

a=4

6 0
3 years ago
Write a polynomial of least degree with roots - 9 and - 2. Write your answer using the variable x and in standard form with a le
ohaa [14]
Ok I will but I don’t know
4 0
3 years ago
Help me please!! I need this for my math test!!
Svetlanka [38]

Answer:

y=+3

Step-by-step explanation:

6 0
3 years ago
Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level. how m
GalinKa [24]

The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

For Addison h=85(1/3)

d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)

Subtracting both distances we get

8\sqrt{2}-6\sqrt{2}  =2\sqrt{2}

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

brainly.com/question/1392973

5 0
3 years ago
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