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SVETLANKA909090 [29]
3 years ago
13

Help quick please give an explanation too.

Mathematics
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer: (x+2)*(x-4)*(x+21) = 0 has "many" solutions

Step-by-step explanation: (x+2) = 0 only has one solution.  

(x+2)*(x-4) = 0 has two solutions, -2 and 4

(x+2)*(x-4)*(x+21) = 0 has three solutions, -2, 4, and -21, etc

If "many" means more than three, then add another (x+XX) term.  A quadratic equation generally has two solutions.

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What is the equation of the line that passes through (4,-1) and (-2,3)
vladimir1956 [14]

Answer: y-6 = 0.5 (x-8)

Step-by-step explanation:

y − y1 = m(x − x1)

The equation is useful when we know:

one point on the line: (x1,y1)

and the slope of the line: m,

and want to find other points on the line.

Have a play with it first (move the point, try different slopes):

8 0
3 years ago
Assume that women's heights are normally distributed with a mean given by , and a standard deviation given by . μ = 63.5 in σ =
tresset_1 [31]
<span> (a) if 1 woman is randomly selected, find the probability that her height is less than 64 in
using z-score formula:
z-score=(x-mu)/sig
(64-63.5)/2.8
=0.18
thus
P(x<64)=P(z<0.18)-=0.5714

B] </span><span> if 33 women are randomly selected, find the probability that they have a mean height less than 64 in
using the central limit theorem of sample means, we shall have:
2.8/</span>√33=0.49
since n>30 we use z-distribtuion
z(64)=(64-63.5)/0.49=1.191
The
P(x_bar<64)=P(x<1.191)=0.8830
8 0
3 years ago
I need help.. i really want to go sleep.. thank you so much...
Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given  \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}

To verify that the above equality is true or false:

Now find \sum\limits_{k=0}^8\frac{1}{k+3}

Expanding the summation we get

\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3} \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

Now find \sum\limits_{i=3}^{11}\frac{1}{i}

Expanding the summation we get

\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

 Comparing the two series  we get,

\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i} so the given equality is true.

2) Given \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}

Verify the above equality is true or false

Now find \sum\limits_{k=0}^4\frac{3k+3}{k+6}

Expanding the summation we get

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}

now find \sum\limits_{i=1}^3\frac{3i}{i+5}

Expanding the summation we get

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}

Comparing the series we get that the given equality is false.

ie, \sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}

6 0
3 years ago
How much ice cream can be put into a cone with radius 3.5 cm and height 12cm​
ivolga24 [154]

Step-by-step explanation:

radius of the cone = 3.5cm

height of cone = 12 cm

volume of cone = pie × r ^2 × h

= 1/3 × 22/7 × 35/10× 35 /10 × 12  = 154 cm^3

the extra icecream cannot be measured untill it's weight is not given.

6 0
3 years ago
Damon says that to write 1 63/100 as a sum of unit fractions, you need 163 addends. Kenya says that you need 64 addends. Who is
stealth61 [152]
Damon is right because the 1 is actually 100/100 so 100 + 63 is 163.
7 0
3 years ago
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