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I am Lyosha [343]
3 years ago
14

Evaluate 10x+1 for x=0

Mathematics
2 answers:
Hatshy [7]3 years ago
8 0

Answer:

1

Step-by-step explanation:

because its 1 10x0+1=1

Anna11 [10]3 years ago
6 0

Answer:1

Step-by-step explanation:

You might be interested in
Why 10^33/2=5*10^32? Explain, pls. :(
german

Answer:

see below

Step-by-step explanation:

10^33/2

Rewriting the numerator as 10 * 10 ^ 32

10 * 10 ^32

--------------------

2

10/2 = 5

5 * 10 ^32

Therefore

10^33/2=5*10^32

4 0
2 years ago
What is 5.37 as a mixed number.
Gnom [1K]
Answer: 5.37 as a mixed number is 5 37/100

To write a decimal number as a mixed number, you must separate the whole part of the decimal part:
5.37 = 5 + 0.37 = 5 0.37
And express the decimal part as a fraction:
0.37=37/100
Then:
Answer: 5.37 as a mixed number is 5 37/100
7 0
3 years ago
if i had 100 cookies and my friend has 50, but he took 50 more from me, and now i have 150 cookies and a dead body how many dead
Svetlanka [38]

Answer:

I know this isn't a real question,but you would have 11 dead bodies,I think

Step-by-step explanation:

8 0
3 years ago
Is the expression 7 (6x-8y) equal to 42x-56y
Bond [772]
Yes☺
you are right 7 times 6 is 42 and 7 times. 8 is 56
3 0
3 years ago
Read 2 more answers
The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
2 years ago
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