Hello!
1. -4x2 + 8x + 32
-4(x2 - 2x - 8)
-4(x2 + 2x - 4x - 8)
-4(x(x + 2) - 4(x + 2))
-4(x + 2)(x - 4)
Answer A.
2. 2x2 + 16x + 30
2(x2 + 8x + 15)
2(x2 + 5x + 3x + 15)
2(x(x + 5) + 3(x + 5))
2(x + 5)(x + 3)
Answer C.
Answer:
(18+21+17+x)÷4=20
x=24 students
Step-by-step explanation:
(18+21+17+x)÷4=20
(56+x)÷4=20
56+x=20*4
56+x=80
x=80-56
x=24
Answer:
p = 25,20 cm
h = 34,32 cm
A(min) = 864,86 cm²
Step-by-step explanation:
Let call p and h dimensions of a poster, ( length , and height respectively) and x and y dimensions of the printed area of the poster then
p = x + 8 and h = y + 12
Printed area = A(p) = 384 cm² and A(p) = x*y ⇒ y = A(p)/x
y = 384 / x
Poster area = A(t) = ( x + 8 ) * ( y + 12 ) ⇒ A(t) = ( x + 8 ) * [( 384/x ) + 12 ]
A(t) = 384 + 12x + 3072/x + 96 A(t) = 480 + 3072/x + 12x
A(t) = [480x + 12x² + 3072 ] / x
A´(t) = [(480 + 24x )* x - (480x + 12x² + 3072]/x²
A´(t) = 0 [(480 + 24x )* x - 480x - 12x² - 3072] =0
480x + 24x² -480x -12x² - 3072 = 0
12x² = 3072 x² = 296
x = 17,20 cm and y = 384/17,20 y = 22,32 cm
Notice if you substitu the value of x = 17,20 in A(t) ; A(t) >0 so we have a minimun at that point
Then dimensions of the poster
p = 17,20 + 8 = 25,20 cm
h = 22.32 + 12 =34,32 cm
A(min) = 25,20 *34.32
A(min) = 864,86 cm²
L: length defective<span>, T </span>: texture defective.
Given that the strip is length defective, the probability that this strip is texture defective is given by
<span><span>P (TL) =P (T ∩ L)=0.008= 0.08.</span><span>P (L)0.1</span><span>| www.imali.info</span></span>
Answer:
<em><u>y = 4 and x = 9</u></em>
Step-by-step explanation:
I hope this is correct
