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3 years ago

What is the solution of the equation?

Mathematics

2 answers:

fredd [130]3 years ago

8 0

Answer: 7

Step-by-step explanation:

$\sqrt{2x-5}+4 = x$

$\sqrt{2x-5} = x-4$ subtracted 4 from both sides

$(\sqrt{2x-5})^2 = (x-4)^2$ squared both sides to eliminate square root

2x - 5 = x² - 8x + 16 expanded right side

0 = x² - 10x + 21 subtracted 2x and added 5 on both sides

0 = (x - 3) (x - 7) factored right side

0 = x - 3 0 = x - 7 applied zero product property

x = 3 x = 7 solved for x

Check:

x = 3

$\sqrt{2(3)-5}+4 = (3)$

$\sqrt{1}+4 = 3$

1 + 4 = 3

FALSE! x = 3 is NOT a valid solution

x = 7

$\sqrt{2(7)-5}+4 = (7)$

$\sqrt{9}+4 = 7$

3 + 4 = 7

TRUE! x = 7 IS a valid solution

Mazyrski [523]3 years ago

6 0

$\sqrt{2x - 5} = x - 4$
$2x - 5 = (x - 4) {}^{2}$
$2x - 5 = x {}^{2} - 8x + 16$
$x {}^{2} - 10x + 21 = 0$
$(x - 3)(x - 7) = 0$
$x = 3 \\ x = 7$

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A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

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Read 2 more answers

Someone please answer and explain this for me!

Nesterboy [21]

Hello!

To find our answer, we will want to multiply our circumference by 15. The reason why is because the circumference is the distance around the circle, or tire in this case. If it goes around 15 times it will have traveled this distance 15 times.

Therefore, we find our circumference, multiplying pi by our diameter. It appears that the total height is 26 in.

26 $\pi$ ≈81.68

Note that it wants us to round for the nearest FOOT. Therefore, we divide our answer by 12 and round to the nearest whole number.

81.68/12≈6.81≈7

Therefore, it will have traveled 7 feet.

I hope this helps!

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