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kvasek [131]
2 years ago
7

Olivia chose 10/21 . How did she get that answer?

Mathematics
1 answer:
Katyanochek1 [597]2 years ago
5 0
Cuz she divided 23 and 8 and got 21 then she added 3+3= 6 + 4= 10
You might be interested in
Simplify 2^4 root 176 + 5^4 root 11
Misha Larkins [42]

Answer:

689 root 11

Step-by-step explanation:

simplify

16x4root11+5^4 root 11

64 root 11 +625 root 11

8 0
2 years ago
The library is 1.75 miles directly north of the school. The park is 0.6 miles directly south of the school. How far is the libra
kumpel [21]
If something is placed the way these places are(directly NORTH and the park is SOUTH) all you have to do is add your two values, lining up your decimal points. 1.75
            +.6
            2.35
Let me know if this helps by making me your brainliest answer
4 0
3 years ago
Read 2 more answers
How to solve this problem?
777dan777 [17]

Answer:

Can not be determined.

Step-by-step explanation:

We can easily notice that the limit is x tends to infinity, whereas x is not present in the given function, we are given (1 + 1/n). So we can not evaluate the given limit for x as parameter, we must have some function of x to solve this problem.

Hence, option C is correct i.e. the limit can not be determined.

5 0
2 years ago
A number represented by x times -0.4 is 2.5, what is the value of x?
-BARSIC- [3]
-0.4x=2.5 divide both sides by -0.4

x=-6.25
5 0
2 years ago
Find the equation of the line which passes through the point
sammy [17]

Answer:

y= \frac{1}{5}x- \frac{56}{5}

Step-by-step explanation:

We want to find the equation of the line that goes through: !6,-10) and is perpendicular to the line

5x + 3y = 2y - 3

Let's simplify the equation of the line to get:

3y - 2y = - 5x + 3

y = - 5x + 3

The slope of this line is -5.

The perpendicular to this line has a slope of

\frac{1}{5}

The equation going through (6,-10) is

y=m(x-x_1)+y_1

We substitute to get;

y= \frac{1}{5} (x-6) - 10

Expand;

y= \frac{1}{5}x- \frac{6}{5}  - 10

y= \frac{1}{5}x- \frac{56}{5}

7 0
2 years ago
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