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tankabanditka [31]
1 year ago
13

Is the U.S. ever on sale to canadians?

Mathematics
1 answer:
Goshia [24]1 year ago
7 0

Answer:

No because Us is powerful country canada is also powerful but U S ever sale to candians

You might be interested in
Find the area of the rhombus in m squared
Nataly [62]

Answer:

\boxed{\bold{100 \ m^2}}

Step By Step Explanation:

Area Of Rhombus: \bold{\frac{pq}{2} }

Identify 'p'

5 + 5 = 10

p = 10

Identify 'q'

10 + 10 = 20

q = 20

\bold{10 \ \cdot\ \ 20 \ = \ 200}

\bold{200 \ \div \ 2 \ = \ 100}

➤ \boxed{\bold{Mordancy}}

4 0
2 years ago
The identification code on a bank card consists of 1 digit followed by 2 letters. The code must meet the following conditions: T
jasenka [17]

Answer:

For the code we have 3 selections.

The first selection is a digit that must be odd, so the options are {1, 3, 5, 7 ,9}

So we have 5 options.

The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)

So here we have 27 - 5 = 22 options

The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.

Here we have 22 - 1 = 21 options.

The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:

C = 5*22*21 = 2310.

There are 2310 different possible codes

8 0
3 years ago
Given a and b are numbers, and a + b = 180, which
Anastasy [175]
If the first one is a = 180 - b, and not 180 - 6, then that is correct.

Also, 360 = 2a + 2b is correct
6 0
2 years ago
the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2
lilavasa [31]

Answer:  c) \dfrac{3}{2} .

Step-by-step explanation:

Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}

Given function : f(x) = x^2

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3

\Rightarrow\ f'(c)=3\ \ \ ...(i)

Since f'(x)=2x

then, at x=c, f'(c)=2c\ \ \ ...(ii)

From (i) and (ii), we have

2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}

Hence, the correct option is c) \dfrac{3}{2} .

4 0
3 years ago
Ilke is deciding whether to take his girlfriend on a dinner date or on a movie date the probability of this having a successful
Lerok [7]

Answer:

The dinner date

Step-by-step explanation:

5 0
3 years ago
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