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Alborosie
2 years ago
6

Use the quadratic formula to find the solution to the question 3x^2-10x+5=0

Mathematics
1 answer:
nlexa [21]2 years ago
3 0

Answer:

x_{1}  = \frac{5 + \sqrt{10} }{3}  and x_{2} = \frac{5 - \sqrt{10} }{3}

Step-by-step explanation:

3x^2-10x+5=0

x_{1} = \frac{10 + \sqrt{10^{2}-4*3*5 } }{2*3}                  x_{2} = \frac{10 - \sqrt{10^{2}-4*3*5 } }{2*3}

x_{1} = \frac{10 + \sqrt{100-60} }{6}                      x_{2} = \frac{10 - \sqrt{100-60} }{6}

x_{1} = \frac{10 + \sqrt{40} }{6}                            x_{2} = \frac{10 - \sqrt{40} }{6}

x_{1} = \frac{10 + 2\sqrt{10} }{6}  / 2                    x_{2} = \frac{10 - 2\sqrt{10} }{6}  / 2

x_{1}  = \frac{5 + \sqrt{10} }{3}                             x_{2} = \frac{5 - \sqrt{10} }{3}

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2 years ago
A and B are independent events. P(A) = 0.50 P(B) = 0.30 What is P(A/B)? a. not enough information. b. 0.30 c. 0.50 d. 0.15
Gala2k [10]

Answer:

0.50

Step-by-step explanation:

P(A|B)=P(A and B)/P(B) is the formula to find a conditional probability when A and B are not independent.

A and B are independent. This means B has no effect of A. This means P(A|B)=P(A).

P(A|B)=P(A) since A and B are independent

P(A|B)=0.50

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2 years ago
Find the cost per roll if a bag of five lunch rolls is $2.50.
Andru [333]
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2 years ago
Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,
torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

y=6x^2+14x

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

\frac{dy}{dx}=12x+14

We know that length of curve

s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx

We have a=-2 and b=1

Using the formula

Length of curve=s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

dt=12dx

dx=\frac{1}{12}dt

Length of curve=s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt

We know that

\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C

By using the formula

Length of curve=s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})

Length of curve=s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})

Length of curve=s=32.66

5 0
2 years ago
Two certificates of deposit pay interest that differ by 3%. Money invested for one year in the first CD earns $240 interest. The
Soloha48 [4]

a = interest rate of first CD

b = interest rate of second CD

and again, let's say the principal invested in each is $X.

\bf a-b=3\qquad \implies \qquad \boxed{b}=3+a~\hfill \begin{cases} \left( \frac{a}{100} \right)X=240\\\\ \left( \frac{b}{100} \right)X=360 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \left( \cfrac{a}{100} \right)X=240\implies X=\cfrac{240}{~~\frac{a}{100}~~}\implies X=\cfrac{24000}{a} \\\\\\ \left( \cfrac{b}{100} \right)X=360\implies X=\cfrac{360}{~~\frac{b}{100}~~}\implies X=\cfrac{36000}{b} \\\\[-0.35em] ~\dotfill\\\\

\bf X=X\qquad thus\qquad \implies \cfrac{24000}{a}=\cfrac{36000}{b}\implies \cfrac{24000}{a}=\cfrac{36000}{\boxed{3+a}} \\\\\\ (3+a)24000=36000a\implies \cfrac{3+a}{a}=\cfrac{36000}{24000}\implies \cfrac{3-a}{a}=\cfrac{3}{2} \\\\\\ 6-2a=3a\implies 6=5a\implies \cfrac{6}{5}=a\implies 1\frac{1}{5}=a\implies \blacktriangleright 1.2 = x\blacktriangleleft

\bf \stackrel{\textit{since we know that}}{b=3+a}\implies b=3+\cfrac{6}{5}\implies b=\cfrac{21}{5}\implies b=4\frac{1}{5}\implies \blacktriangleright b=4.2 \blacktriangleleft

3 0
3 years ago
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