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3 years ago

9

How do i solve this plss helpp

2 answers:

kkurt [141]3 years ago

4 0

The answer is -12!
Explanation: (view image)

maria [59]3 years ago

3 0

Hope this helps!! :)))))))

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Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

2 years ago

Find the missing sides, and give the simplified form of the square root

Brilliant_brown [7]

Answer:

in the pics

Step-by-step explanation:

In case you are wondering, I used the pythagorean theorem. If you want to learn how to use it I suggest you google it, there are plenty of sites that explain how to use it.

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below :)

7 0

3 years ago

Please help no files

valentina_108 [34]

Answer:

x =1 for the first problem

and for second question

Step-by-step explanation:

Let's solve your inequality step-by-step.

|2x+5|>0.5

Solve Absolute Value.

|2x+5|>0.5

We know either2x+5>0.5or2x+5<−0.5

2x+5>0.5(Possibility 1)

2x+5−5>0.5−5(Subtract 5 from both sides)

2x>−4.5

2x2>−4.52(Divide both sides by 2)

x>−2.25

2x+5<−0.5(Possibility 2)

2x+5−5<−0.5−5(Subtract 5 from both sides)

2x<−5.5

2x2<−5.52(Divide both sides by 2)

x<−2.75

Answer:

x>−2.25 or x<−2.75

3 0

3 years ago

What diagonals does a kite intersect at?

Romashka [77]

The diagonals of a kite meet at a right angle.

3 0

3 years ago

Solve by completing the square: 6x^2+39=5+12x

Marta_Voda [28]

6x^2-12x=-34
x^2-2x=-34/6
x^2-2x=-17/3

x^2-2x+1= -17/3+1/1
x^2-2x+1= -14/3

(x-1)^2= -14/3
x-1= √-14/3

Final answer: No solution (you cannot find the root of a negative number)

8 0

3 years ago