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Over [174]
3 years ago
13

What is the number of degrees of rotational symmetry for a parallelogram?​

Mathematics
1 answer:
In-s [12.5K]3 years ago
5 0

Answer:

A parallelogram has 180 degrees of rotational symmetry

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A) Work out the value of 213 + (29)2
Alina [70]

a) 2^13 / 2^18

b) 216

6 0
3 years ago
X/2 -5=20 please give details
vladimir1956 [14]

Answer: x=50

how?

\frac{x}{2} - 5= 20 (first ADD 5 with 20 whatever you do on one side you have to do it to the other) so that gives us:

\frac{x}{2}=25 (we see that 2 is being divided by x and we want X alone so the inverse of division is MULTIPLICATION)

25 x 2= 50

therefor the x=50

6 0
3 years ago
Which can be used to expand the expression below?
KiRa [710]

Answer:

Distributive Property

Step-by-step explanation:

We have

5(3x - 6/7)

Distributive Property would work because 5 has to be mutliplied by 3x and -6/7 to rid the parenthesis and also give a final answer for the equation. Any others will lead to a wrong answer.

7 0
3 years ago
Read 2 more answers
Use the definition of continuity to determine whether f is continuous at a.
dmitriy555 [2]
f(x) will be continuous at x=a=7 if
(i) \displaystyle\lim_{x\to7}f(x) exists,
(ii) f(7) exists, and
(iii) \displaystyle\lim_{x\to7}f(x)=f(7).

The second condition is immediate, since f(7)=8918 has a finite value. The other two conditions can be established by proving that the limit of the function as x\to7 is indeed the value of f(7). That is, we must prove that for any \varepsilon>0, we can find \delta>0 such that

|x-7|

Now,


|f(x)-f(7)|=|5x^4-9x^3+x-8925|

Notice that when x=7, we have 5x^4-9x^3+x-8925=0. By the polynomial remainder theorem, we know that x-7 is then a factor of this polynomial. Indeed, we can write

|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|

This is the quantity that we do not want exceeding \varepsilon. Suppose we focus our attention on small values \delta. For instance, say we restrict \delta to be no larger than 1, i.e. \delta\le1. Under this condition, we have

|x-7|

Now, by the triangle inequality,


|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275

If |x|, then this quantity is moreover bounded such that

|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955

To recap, fixing \delta\le1 would force |x|, which makes


|x-7||5x^3+26x^2+182x+1275|

and we want this quantity to be smaller than \varepsilon, so


6955|x-7|

which suggests that we could set \delta=\dfrac{\varepsilon}{6955}. But if \varepsilon is given such that the above inequality fails for \delta=\dfrac{\varepsilon}{6955}, then we can always fall back on \delta=1, for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set \delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}.

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that f(x) is indeed continuous at x=7.
5 0
3 years ago
HELP PLEASE!!!!!!!!! ASAP
beks73 [17]

Answer:

7.5

Step-by-step explanation:

Because XY = XW, by the property of an Isosceles triangle.

If you find it is helpful, plz give me Brainliest.

5 0
3 years ago
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