For this case we have a function of the form:
y = f (x)
Where,
x: independent variable
y: dependent variable
To answer the question, we must see in the table the values of the independent variable x, for which the values of the dependent variable f (x) are negative.
We have then that the interval that fulfills this condition is from minus infinity to minus 4 without including the 4.
Thus,
(–∞, –4)
Answer:
The entire interval over which the function, f(x), is negative is:
D) (–∞, –4)
<span>F for Frank, A or Alice.
F(initial)=1.95 inches
A(initial)=1.50 inches
Frank's equation at .25 inches per year and t representing year variable.
F=1.95+.25t
Alice's equation at .40 inches per year and t representing year variable.
A=1.5+.40t
To figure out how old they will be when their beaks are the same lengths set the equations equal to eachother as the equations are length.
1.95+.25t=1.5+.40t
.45=.15t
t=3 years</span>
The answer is <span>y= (x-3)^2 + 9
I hope this helps</span>
Penjelasan langkah demi langkah:
1)
![= 243^{\frac{2}{3} }\\= (\sqrt[3]{243})^2\\= 7^2\\= 49](https://tex.z-dn.net/?f=%3D%20243%5E%7B%5Cfrac%7B2%7D%7B3%7D%20%7D%5C%5C%3D%20%28%5Csqrt%5B3%5D%7B243%7D%29%5E2%5C%5C%3D%207%5E2%5C%5C%3D%2049)
2) √32 +3√18-2√50
= √16*2 +3√9*2-2√25*2
= 4√2 + 3(3√2)-2(5√2)
= 4√2 + 9√2-10√2
= 13√2-10√2
= 3√2
3) 1000 ⅔×64⅙
![= (\sqrt[3]{1000}) ^2 \times (2^6)^{1/6} \\= 10^2 \times 2\\= 100 \times 2\\= 200](https://tex.z-dn.net/?f=%3D%20%28%5Csqrt%5B3%5D%7B1000%7D%29%20%5E2%20%5Ctimes%20%282%5E6%29%5E%7B1%2F6%7D%20%20%5C%5C%3D%2010%5E2%20%5Ctimes%202%5C%5C%3D%20100%20%5Ctimes%202%5C%5C%3D%20200)
4) 3/4+√2
5) 2√3×√18
= 2√3×√9*2
= 2√3×3√2
= (2*3)(√3*√2)
= 6√6
6) 12/3+√3
= 4+√3
7) √1000—2√40
= 10 -2 (√4*10)
= 10-2(2√10)
= 10 - 4√10
8) 2- ¹+3-¹
9)
Jika pernyataannya opsional, penyebutnya adalah 1
10) 2√3×√18
= 2√3×√9*2
= 2√3×3√2
= (2*3)(√3*√2)
= 6√6
Answer:
I believe the answer is the 2nd box{y-2=1/6 (x+10)}, 3rd box{y-1=1/6(x+4)}, and the 5th box{y=1 x/6 +1/3}.
Step-by-step explanation:
