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Kazeer [188]
3 years ago
10

If a zero is _, then the graph of its function only touches the x–axis at that zero. If a zero is _, then the graph of its funct

ion crosses the x–axis at that zero.

Mathematics
2 answers:
SOVA2 [1]3 years ago
3 0

Answer:

even then odd

Step-by-step explanation:

myrzilka [38]3 years ago
3 0

Answer:

even

odd

Step-by-step explanation:

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3 years ago
Sarah is carrying out a series of experiments which involve using mcreasing amounts of a chemical. In the
Marianna [84]

Sarah is carrying out a series of experiments which involve using increasing amounts of a chemical. In the  first experiment she uses 6g of the chemical and in the second experiment she uses 7.8 g of the chemical

(i)Given that the amounts of the chemical used form an arithmetic progression find the total amount of  chemical used in the first 30 experiments

(ii)Instead it is given that the amounts of the chemical used for a geometric progression. Sarah has a  total of 1800 g of the chemical available. Show that the greatest number of experiments possible satisfies the inequality: 1.3^N \leq 91 and use logarithms to calculate the value of N.

Answer:

(a)963 grams

(b)N=17

Step-by-step explanation:

(a)

In the first experiment, Sarah uses 6g of the chemical

In the second experiment, Sarah uses 7.8g of the chemical

If this forms an arithmetic progression:

First term, a =6g

Common difference. d= 7.8 -6 =1.8 g

Therefore:

Total Amount of  chemical used in the first 30 experiments

S_n=\dfrac{n}{2}[2a+(n-1)d] \\S_{30}=\dfrac{30}{2}[2*6+(30-1)1.8] \\=15[12+29*1.8]\\=15[12+52.2]\\=15*64.2\\=963$ grams

Sarah uses 963 grams in the first 30 experiments.

(b) If the increase is geometric

First Term, a=6g

Common ratio, r =7.8/6 =1.3

Sarah has a total of 1800 g

Therefore:

Sum of a geometric sequence

S_n=\dfrac{a(r^N-1)}{r-1} \\1800=\dfrac{6(1.3^N-1)}{1.3-1} \\1800=\dfrac{6(1.3^N-1)}{0.3}\\$Cross multiply\\1800*0.3=6(1.3^N-1)\\6(1.3^N-1)=540\\1.3^N-1=540\div 6\\1.3^N-1=90\\1.3^N=90+1\\1.3^N=91

Therefore, the greatest possible number of experiments satisfies the inequality

1.3^N \leq 91

Next, we solve for N

Changing 1.3^N \leq 91 to logarithm form, we obtain:

N \leq log_{1.3}91\\N \leq \dfrac{log 91}{log 1.3}\\ N \leq  17.19

Therefore, the number of possible experiments, N=17

3 0
3 years ago
Suppose f(x)=x^3 find the graph of f(x+1)
Basile [38]
F(x+1) = (x + 1)^3
= 3x + 3
= 3(0) + 3
= 3


I think is 3
5 0
3 years ago
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