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Fudgin [204]
2 years ago
12

In ARST, segment RS = segmentTR and mZT=55. Find mZS. * ????

Mathematics
1 answer:
Karo-lina-s [1.5K]2 years ago
8 0

Answer:

The answer would be 60.

Step-by-step explanation:

The four students in the table below each recorded the time and distance traveled while exercising. exercising distance (miles) time (minutes) gia 2 30 harris 5 50 ian 3 40 jackson 4 80 which list ranks the students from fastest walker to slowest walker? jackson, gia, ian, harris harris, jackson, ian, gia harris, ian, gia, jackson jackson, harris, ian, gia.Which of these triangle pairs can be mapped to each other using a single translation? cof hn

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a motorboat maintained a constant speed of 18 miles per hour relative to the water and going 35 miles upstream and then return.
AveGali [126]
<span>150+10c+150-10c=1.5(15-c)(15+c)300=1.5(225-c^2)300=337.5-1.5c^2200=225-c^2c^2=25c=5speed of current=5 mph<span>
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6 0
3 years ago
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You begin solving the equation 3+4x=513+4x=51 by subtracting 33 from both sides. Which is the best choice for Step 22?
bonufazy [111]
33 +4x = 513
4x = 480 (x = 480/4: x = 120)


Answer
Step 2 Divide each side by 4
7 0
3 years ago
Please help. Also please show your work. Y/5 - 2/5
Maurinko [17]

Answer:

0.2(y-2)

Step-by-step explanation:

factor out 1/5 from the expression

1/5×(y-2)

0.2(y-2)

8 0
2 years ago
Please help if you are good at math :( thank you
ValentinkaMS [17]

Answer:

your answer will be

A)(6,10)

B)(2,12)

C)(4,4)

Step-by-step explanation:

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7 0
3 years ago
A growth medium is inoculated with 1,000 bacteria, which grow at a rate of 15% each day. What is the population of the culture 6
ddd [48]

Answer:

The population of bacteria after 6 days is 2,313.06

Step-by-step explanation:

Given as :

The initial population of bacteria = i = 1,000 bacteria

The growth rate of bacteria per day = 15%

Let The population of bacteria after 6 days = f

The time period of growth = 6 days

<u>Now, According to question</u>

The population of bacteria after 6 days = initial population × (1+\dfrac{\textrm rate}{100})^{\textrm time}

Or, f = i × (1+\dfrac{\textrm r}{100})^{\textrm t}

Or, f = 1000 × (1+\dfrac{\textrm 15}{100})^{\textrm 6}

Or, f = 1000 × (1.15)^{6}

Or, f = 1000 × 2.31306

∴  f = 2,313.06

So,The population of bacteria after 6 days = f = 2,313.06

Hence,The population of bacteria after 6 days is 2,313.06  Answer

3 0
3 years ago
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