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4 years ago

Which is a solution to the equation? (x - 3)(x - 5) = 35 x= -8 x=-5 x=2 x = 10

Mathematics

2 answers:

JulijaS [17]4 years ago

4 0

(x-3)(x-5)=35

Multiply the brackets together

x*x-5*x-3*x-3*5

x^2-8x+15=35

Move +35 to the other side. Sign changes from +35 to -35

x^2-8x+15-35=35-35

x^2-8x-20=0

(x-10)(x+2)=0

x-10=0 x+2=0

Move -10 to the other side. Sign changes from -10 to +10

x-10+10=0+10 x+2-2=0-2

x=10 x=-2

Answer: x=10

olga2289 [7]4 years ago

3 0

Answer: x = 10

Just plug in the different numbers to see which ones equal 35.

I’m going to plug in x = -8 into the equation
(x - 3)(x - 5) = 35
(-8 - 3)(-8 - 5) = 35 (substitute -8 for x)
(-11)(-13) = 35
143 = 35
This answer is wrong because 143 does not equal 35

Now let’s plug in x = 10
(x - 3)(x - 5) = 35
(10 - 3)(10 - 5) = 35 (substitute 10 for x)
(7)(5) = 35
35 = 35
This solution works because 35 equals 35

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A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) Large and small drains are opened

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) Only the small drain is opened

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$