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Y_Kistochka [10]

3 years ago

12

Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.

1 answer:

xxTIMURxx [149]3 years ago

6 0

The area of the triangle is (1/2)(DE)^2*tan(60°) = (1/2)*2*√3 = √3.

Selection C is appropriate.

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Solve using quadratic formula for 2x^(2)-10x+5=0

sergij07 [2.7K]

Answer:

$x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}$

Step-by-step explanation:

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-10,\:c=5$

$x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}$

$\sqrt{\left(-10\right)^2-4\cdot \:2\cdot \:5}$

Apply exponent rule: (-a)^n=a^n, if n is even

$\left(-10\right)^2=10^2$

$=\sqrt{10^2-4\cdot \:2\cdot \:5}$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:2\cdot \:5=40$

$=\sqrt{10^2-40}$

$10^2=100$

$=\sqrt{100-40}$

$\mathrm{Subtract\:the\:numbers:}\:100-40=60$

$=\sqrt{60}$

$60\:\mathrm{divides\:by}\:2\quad \:60=30\cdot \:2$

$=2\cdot \:30$

$30\:\mathrm{divides\:by}\:2\quad \:30=15\cdot \:2$

$=2\cdot \:2\cdot \:15$

$15\:\mathrm{divides\:by}\:3\quad \:15=5\cdot \:3$

$=2\cdot \:2\cdot \:3\cdot \:5$

$2,\:3,\:5\mathrm{\:are\:all\:prime\:numbers,\:therefore\:no\:further\:factorization\:is\:possible}$

$=2\cdot \:2\cdot \:3\cdot \:5$

$=2^2\cdot \:3\cdot \:5$

$=\sqrt{2^2\cdot \:3\cdot \:5}$

Apply Radical Rule:

$=\sqrt{2^2}\sqrt{3\cdot \:5}$

Apply Radical Rule:

$\sqrt{2^2}=2$

$=2\sqrt{3\cdot \:5}$

$\mathrm{Refine}$

$=2\sqrt{15}$

$x_{1,\:2}=\frac{-\left(-10\right)\pm \:2\sqrt{15}}{2\cdot \:2}$

$\mathrm{Separate\:the\:solutions}$

$x_1=\frac{-\left(-10\right)+2\sqrt{15}}{2\cdot \:2},\:x_2=\frac{-\left(-10\right)-2\sqrt{15}}{2\cdot \:2}$

$\frac{-\left(-10\right)+2\sqrt{15}}{2\cdot \:2}$

Apply Rule -(-a)=a

$=\frac{10+2\sqrt{15}}{2\cdot \:2}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4$

$=\frac{10+2\sqrt{15}}{4}$

$=\frac{2\left(5+\sqrt{15}\right)}{4}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=\frac{5+\sqrt{15}}{2}$

$\frac{-\left(-10\right)-2\sqrt{15}}{2\cdot \:2}$

$=\frac{10-2\sqrt{15}}{2\cdot \:2}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4$

$=\frac{10-2\sqrt{15}}{4}$

$=\frac{2\left(5-\sqrt{15}\right)}{4}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=\frac{5-\sqrt{15}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}$

8 0

2 years ago

What is a sum for 4/3.

Margarita [4]

A sum is addition. This problem is division... But 4/3= 1.33333.

4 0

2 years ago

PLEASE HELP I NEED THE ANSWER NO LINKS OR WEBSITES ANSWER NOW What is the zero of the linear function graphed below?

aniked [119]

Answer:

0 of a polynomial is that point where the graph cuts the x axis so the answer is

<h2>-2</h2>

4 0

2 years ago

Does anyone know???!

jok3333 [9.3K]

Answer:

#8 might be y= -3x + 5

Step-by-step explanation:

This may help

5 0

2 years ago

Read 2 more answers

If a quadratic function has x-intercepts at (-7,0) and 10,0), where would

madam [21]

Answer:

x = 1.5

Step-by-step explanation:

Since the axis of symmetry is between the 2 x intercepts, we can find it by finding the average of the x-coordinates of the 2 points.

(-7 + 10) / 2

3/2

= 1.5

So, the axis of symmetry is at x = 1.5

8 0

2 years ago