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yaroslaw [1]
3 years ago
9

What is 8 min 32 sec + 37 min 18 sec + 15 sec

Mathematics
2 answers:
Alex_Xolod [135]3 years ago
7 0
Answer: Pretty sure it’s 46 minutes and 5 seconds.
Bumek [7]3 years ago
5 0

I belive it is

18 sec + 32 sec +15 sec = 65 sec (1 min 5 sec)

1 min + 8 min + 37 min= 46 min

46 mins 5 secs


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Solve for x in this inequality. Show all your work and<br> graph your solution.<br> 4(x-3)&lt;2x-4
maw [93]

Answer:

x<4

Step-by-step explanation:

To solve for x in the inequality equation, the terms should be rearranged such that x is on one side and integers are on the other side.

First, expand 4(x-3).

This will obtain:

4x - 12 < 2x - 4

Now, shift x terms and integers on each side.

4x - 2x <  - 4 + 12

After simplifying, it will get:

2x < 8

Finally, we can solve for x.

x < 4

We can also draw the respective graph (please don't mind my drawings), where the area shaded in green is the range.

6 0
4 years ago
Hey guys,<br><br> Do you guys know the answer to this TTM problem?<br><br> Thank You!
tensa zangetsu [6.8K]
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4 0
3 years ago
How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between
-BARSIC- [3]
The Fundamental Theorem of Calculus regarding geometry states that 
\int\limits^b_a g{(x)} \, dx = F(b)-F(a)

Where F is the indefinite integral of g(x)

The first step is to integrate g(x)
\int\ {4x} \, dx = \frac{4x^{1+1} }{1+1} = \frac{4x^{2} }{2} =2 x^{2}

Then substitute the value of b and a=1 into 2x^{2}

[2 (b)^{2}]-[2 (1)^{2}] = 240
2b^{2} -2=240
2b^{2}=240+2
2b^{2}=242
b^{2}= \frac{242}{2}
b^{2}=121
b=11

Hence the limit of the area under g(x) is between a=1 and b=11

6 0
3 years ago
Y &gt; 3x +2<br><br>I'm having problems solving Inequalities.<br>Can someone help me????
melomori [17]
Sure! What other examples do you have?
6 0
3 years ago
Hey guys I’m a little stuck ?
Rashid [163]
Tbh Ita hard to under
7 0
3 years ago
Read 2 more answers
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