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sergey [27]
3 years ago
14

2.Solve x squared - 4x by Factoring

Mathematics
1 answer:
777dan777 [17]3 years ago
8 0

Answer:

Step-by-step explanation:

hello:

x²-4x=0  means : x(x-4)=0

x=0 or x-4=0

conclusion : x=0 and x= 4   ...answer : B

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31-18÷6 how would you solve this using PEMDAS?
sergeinik [125]
P: Parenthesis
E: Equation 
M:Multiplication 
D: Division 
A : addition 
S: Subtraction

So first you need to divide because it is before so you divide 18 by 6 which is 3 then you subtract (31 - 3) which gives you 28
answer : 28

7 0
3 years ago
Read 2 more answers
For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?
Bond [772]
Given that r=\theta, then r'=1

The slope of a tangent line in the polar coordinate is given by:

m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}

Thus, we have:

m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}



Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

\sin\theta+\theta\cos\theta=0 \\  \\ \theta\cos\theta=-\sin\theta \\  \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488



Part B:

For vertical tangent lines, \frac{1}{m} =0

Thus, we have:

\cos\theta-\theta\sin\theta=0 \\  \\ \Rightarrow\theta\sin\theta=\cos\theta \\  \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>
3 0
3 years ago
Pls solve for x!: 2x−5.6 over 3 = 1−x over 1.5
olga2289 [7]

Answer:

Step-by-step explanation:

1 2/3 (5 2/7)

5 0
3 years ago
A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.41 mpa and a sample standa
nignag [31]

For this problem, the confidence interval is the one we are looking for. Since the confidence level is not given, we assume that it is 95%.

 

The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n


Where:

<span>
</span>

α= 5%


α/2 = 2.5%


t 0.025, 19 = 2.093 (check t table)


n = 20


df = n – 1 = 20 – 1 = 19

So plugging in our values:


8.41 ± 2.093 * 0.77 √ 1 + 1/20


= 8.41 ± 2.093 * 0.77 (1.0247)


= 8.41 ± 2.093 * 0.789019


= 8.41 ± 1.65141676


<span>= 6.7586 < x < 10.0614</span>

3 0
2 years ago
Plz help....according to the graph
Levart [38]
Hi the answer to this is 

 A - Vertical Asymptote
7 0
3 years ago
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