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Volgvan
3 years ago
6

5a+5-3a=13 i needed to make it 20 charatcers

Mathematics
2 answers:
defon3 years ago
7 0

Answer:

5a plus 5 minus 3a is equal 13

Step-by-step explanation:

melisa1 [442]3 years ago
7 0

Answer:

a=4

Step-by-step explanation:

5a+5-3a=13

Subtract 5 from both sides

5a-3a=8

2a=8

Divide both sides by 2

4 5a+5-3a=13\\    -5       -5 \\ \\5a-3a=8\\2a=8\\\frac{2a}{2} =\frac{8}{2\\} \\a=4

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Carry out three steps of the Bisection Method for f(x)=3x−x4 as follows: (a) Show that f(x) has a zero in [1,2]. (b) Determine w
ELEN [110]

Answer:

a) There's a zero between [1,2]

b) There's a zero between [1.5,2]

c) There's a zero between  [1.5,1.75].

Step-by-step explanation:

We have f(x)=3^x-x^4

A)We need to show that f(x) has a zero in the interval [1, 2]. We have to see if the function f is continuous with f(1) and f(2).

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(2)=3^2-(2)^4=9-16=(-7)

We can see that f(1) and f(2) have opposite signs. And f(1)>f(2) and the function is continuous, this means that exists a real number c between the interval [1,2] where f(c)=0.

B)We have to repeat the same steps of A)

For the subinterval [1,1.5]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1) and f(1.5) have the same signs, this means there's no zero in the subinterval [1,1.5].

For the subinterval [1.5,2]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.5) and f(2) have opposite signs, this means there's a zero between the subinterval [1.5,2].

C)We have to repeat the same steps of A)

For the subinterval [1,1.25]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5

f(1) and f(1.25) have the same signs, this means there's no zero in the subinterval [1,1.25].

For the subinterval [1.25,1.5]:

f(x)=3^x-x^4\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1.25) and f(1.5) have the same signs, this means there's no zero in the subinterval [1.25,1.5].

For the subinterval [1.5,1.75]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)

f(1.5) and f(1.75) have opposite signs, this means there's a zero between the subinterval [1.5,1.75].

For the subinterval [1.75,2]:

f(x)=3^x-x^4\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.75) and f(2) have the same signs, this means there isn't a zero between the subinterval [1.75,2].

The graph of the function shows that the answers are correct.

6 0
3 years ago
Please I need help!!
Ne4ueva [31]

Answer:

From my thoughts I think the answer is 28

7 0
2 years ago
What is the Estimate of 3583.37
Softa [21]

Answer: 3583.37 ≈ 3500

Step-by-step explanation:

5 0
3 years ago
if it takes 3 cups of sugar and 4 1/2 cups of flour for the recipe how many cups of flour for 4 cups of sugar
Vitek1552 [10]

Answer:6 cups

Step-by-step explanation:

3/4.5=4/x

Cross product

3x=18

X=18/3

X=6

8 0
3 years ago
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

              Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
3 years ago
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