Answer:324
Step-by-step explanation:
int i = 42.7; /* konwersja z double do int */
float f = i; /* konwersja z int do float */
double d = f; /* konwersja z float do double */
unsigned u = i; /* konwersja z int do unsigned int */
f = 4.2; /* konwersja z double do float */
i = d; /* konwersja z double do int */
char *str = "foo"; /* konwersja z const char* do char* [1] */
const char *cstr = str; /* konwersja z char* do const char* */
void *ptr = str; /* konwersja z char* do void* */
Podcza
The answer should be 6f. You multiply the 3 and the 2 and then put the f at the end (:
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>
Since you're only given the one vector, I assume you're supposed to find the magnitude of <em>v</em>, or ||<em>v</em>||, using the dot product.
We have
<em>v</em> • <em>v</em> = (-2, -1) • (-2, -1) = (-2)² + (-1)² = 5
so that
||<em>v</em>|| = √(<em>v</em> • <em>v</em>) = √(5)
and the answer is C.