Hey Ive taken this assesment. The answer is 55
Hey there
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The correct answer is:
f(x+5)=(x+5)^2+3(x+5)-10
f(x+5)=x^2+10x+25+3x+15-10
f(x+5)=x^2+13x+30 and f(x+5)=x^2+kx+30 so
k=13
Now factor x^2+13x+30
Find j and k such that jk=ac=30 and j+k+b=13 so j and k are 10 and 3 so
(x+3)(x+10)
So the two zeros occur when x=-3 and -10 the smallest of which is:
x=-10
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Hope this helps you
Answer:
Possible dimensions are 16 × 4 inches and 32 × 2 inches.
Step-by-step explanation:
We are given that,
Area of the frame = 64 inch² = Length × Width
Now, we have that 64 can be factored as,
64 = 32 × 2
64 = 16 × 4
64 = 8 × 8
Since, the frame is in a rectangular shape. So, we reject the possibility of 8 × 8 inches frame.
Thus, the possible dimensions are,
Set 1: 16 × 4 inches
Set 2: 32 × 2 inches.
Answer:
2 2/5
Step-by-step explanation:
I'm not smart but you can trust me :)
