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2 years ago

8

Use the quadratic formula to solve the equation 25x^2-30x+25=0

1 answer:

Reika [66]2 years ago

8 0

Answer: $x=\frac{3}{5}+i\frac{4}{5},\:x=\frac{3}{5}-i\frac{4}{5}$

Step-by-step explanation:

$25x^2-30x+25=0$

$x_{1,\:2}=\frac{-\left(-30\right)\pm \sqrt{\left(-30\right)^2-4\cdot \:25\cdot \:25}}{2\cdot \:25}$

$\sqrt{\left(-30\right)^2-4\cdot \:25\cdot \:25}$

$=\sqrt{30^2-4\cdot \:25\cdot \:25}$

$=\sqrt{30^2-2500}$

$=i\sqrt{2500-30^2}$

$=40i$

$x_{1,\:2}=\frac{-\left(-30\right)\pm \:40i}{2\cdot \:25}$

$x_1=\frac{-\left(-30\right)+40i}{2\cdot \:25},\:x_2=\frac{-\left(-30\right)-40i}{2\cdot \:25}$

$x=\frac{3}{5}+i\frac{4}{5},\:x=\frac{3}{5}-i\frac{4}{5}$

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Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

<u>iii) Add the volume of the two frustrums</u>

$Volume=27.6m^3+38.3m^3=65.9m^3$

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Please help i will give branliest!!!

Pepsi [2]

Answer:b

Step-by-step explanation:

3.5*5*.5=8.75

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