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astra-53 [7]
2 years ago
5

Solve for z 7z < 21 z=

Mathematics
1 answer:
Step2247 [10]2 years ago
3 0

Answer:

z=5

Step-by-step explanation:

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HELP WILL MARK YOU BRAINLIEST HURRY IM ON A TIME LIMIT!’
Keith_Richards [23]

(3x+5)+(x+3)=180

4x+8=180

4x=172

x=43

3(43)+5

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6 0
2 years ago
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What is 0.123 rounded to the nearest hundredth
Sholpan [36]
You would round to 0.12, What you would do is look at the placement of each number. The 1 is in the tenths place, the 2 is in the hundreds and the 3 is in the thousands. So you look at the three and see if you would round up or down, if it is 5 or greater you round up if it is 4 or less then round down. This is less so you would round down to 0.12
4 0
3 years ago
]FIRST PERSON THAT ANSWERS CORRECTLY WILL GET BRAINLIST AND 80 POINTS!!!!
olchik [2.2K]

Answer:

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3 0
2 years ago
Jill bought a car for $50,000. Her car depreciates at a rate of 10% a year. create an equation for the problem
Vadim26 [7]

Answer:

The value of car after n years at the depreciation rate is                              $ 50,000 (0.9)^{n}  .

Step-by-step explanation:

Given as :

The cost of the car that Jill bought = $ 50,000

The depreciation rate of car value = r = 10 % a years

Let The car after n years of depreciation = $ A

Now, According to question

The cost of car after n years of depreciation = initial cost of car × (1 - \dfrac{\textrm rate}{100})^{\textrm time}

Or, $ A = $50,000 × (1 - \dfrac{\textrm r}{100})^{\textrm n}

Or, $ A = $50,000 × (1 - \dfrac{\textrm 10}{100})^{\textrm n}

Or, $ A = $50,000 × (\frac{90}{100})^{n}

I.e $ A = $50,000 × (0.9)^{n}

So, value of car after n years = $ 50,000 (0.9)^{n}

Hence The value of car after n years at the depreciation rate is                      $ 50,000 (0.9)^{n}  . Answer

6 0
2 years ago
The price of a brick today is 49¢. This is 3¢ less than
Luda [366]

Answer:

13 cents

Step-by-step explanation:

Let x be the price 20 years ago.

.49 = 4x - .03

.52 = 4x

.52/4 = x

x = .13

7 0
3 years ago
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