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Delicious77 [7]

3 years ago

14

Math questions plz help 10-11b

1 answer:

evablogger [386]3 years ago

6 0

Question number 10 is the same as the previous questions add them up and make them equal to 180 degrees, and then solve for the x variable.

For the second part of the question, simply plug in the value of x for each expression for R and S.

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A new truck can travel 350 miles on 25 gallons of gas. How far can it travel on 60 gallons of gas? What is the car's gas milage?

nataly862011 [7]

ANSWER:

The MPG is 1/14.
The car can travel 840 miles with 60 gal. of gas.

4 0

2 years ago

Convert a mark of 72 out of 120 to a mark out of 300

Semmy [17]

Answer:

180 out of 300

Step-by-step explanation:

Basically,

300÷120=2.5

Thereof,

72×2.5=180.

<em>Hope this helps!</em>

<em>tag </em><em>brainliest</em><em> </em><em>please.</em><em>.</em><em>.</em>

5 0

3 years ago

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

3 years ago

Please please please help me

alexira [117]

A and can I be brainliest? :3

7 0

3 years ago

Please help me.Find the value of x in the isosceles triangle shown below

blagie [28]

There is an isosceles triangle.

The triangle is like two right triangles put together.

To find out the height of the triangle, we remove the other right triangle and we use the other triangle to find the height using the Pythagorean theorem.

Phytagoras Theorem :

b² = c² - a²

x² = 8² - (6÷2)²

x² = 64 - 3²

x² = 64 - 9

x² = 55

x = √55

So, the value of x is √55 (B)

<em>Hope </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>

4 0

3 years ago