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IgorC [24]
3 years ago
9

Help me on my Homework it is not all of it but I can do the rest myself.

Mathematics
2 answers:
n200080 [17]3 years ago
8 0
The person above me is right <3
Rudik [331]3 years ago
4 0

just label the first quartile, median, third quartile, and maximum

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Solve each equation. 6n+11=-13 and -4(x-5)=8
Luba_88 [7]

Answer:

n=-4     x=-7

Step-by-step explanation:

6 0
4 years ago
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What is <br>1.63 x 0.98 =​
yarga [219]

Answer:

1.5974

Step-by-step explanation:

8 0
3 years ago
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you made a batch of 24 cookies for school your brother ate 7 of them without asking you. What fraction of your cookies are left?
polet [3.4K]
17/24.

Explanation:
24/24 is your original amount.
24 - 7 = 17.
So now you have 17/24! :D
4 0
3 years ago
A car starts with a dull tank of gas. After driving around 5he city, 1/7 of the gas has been used. With the rest of the gas in t
Sati [7]

Answer:

\frac{2}{7}

Step-by-step explanation:

Given:

A car starts with a dull tank of gas

1/7 of the gas has been used around the city.

With the rest of the gas in the car, the car can travel to and from Ottawa three times.

Question asked:

What fractions of a tank of gas does each complete trip to Ottawa use?

Solution:

Fuel used around the city = \frac{1}{7}

Remaining fuel after driving around the city = 1 - \frac{1}{7}

                                                                         =    \frac{7 - 1}{7}  = \frac{6}{7}

According to question:

As from the rest of the gas in the car that is \frac{6}{7}, the car can complete 3 trip to Ottawa  which means,

By unitary method:

The car can complete 3 trip by using = \frac{6}{7} tank of gas.

The car can complete 1 trip by using =  \frac{6}{7} \div 3

                                                             =\frac{6}{7} \times\frac{1}{3}

                                                             =  \frac{6}{21}

                                                             = \frac{2}{7} tank of gas

Thus, \frac{2}{7} tank of gas used for each complete trip to Ottawa.

5 0
3 years ago
Value of x in log5x = 4logx5 is
olchik [2.2K]
Log5 x=4 logx 5
Ln x/ln5=4(ln5 / ln x)                                 logx z=ln x / ln z
least common multiple=ln5 * ln x

ln²x=4ln²5
√(ln² x)=√(4 ln²5)
ln x=2ln 5
x=e^2ln5=25

Answer: x=25.

4 0
3 years ago
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