Please help me with this I will give brainliest

I need help with both of them please!( Sample questions 13 and 14)

zaharov [31]

$First~exercise! \\ We~ have `there ~some~ triangles.(2) \\ 1)~In~ a ~triangle~,the~degree~of~all~angles~is~180! \\ We~have~2~angles~known,so: \\ 180=x+45+56 \\ 180=101+x \\ x=79 \\ \\ Second~triangle! \\ We~have~to~find~2~angles! \\ An ~elongated`~angle~have~180~degrees! \\ 180=79+50+y \\ y=51. \\ In~the~second~triangle~,we~have~to~find~the`third~angle. \\ We~have~a~right~angle,so--\ \textgreater \ 90~degrees! \\ 180=90+51+z \\ z=39! \\ We~found~all~the~angles!(We~are`happy~now!!!~:))) \\ Second~exercise! \\ With~the~Pythagoras' Theorem,we~can~find~the~diagonal~in~this~ \\ rectangle! \\ c ^{2} =10 ^{2} +14 ^{2} \\ c ^{2} =100+196 \\ c= \sqrt{296} \\ c=17,20...$

8 0

4 years ago

How to find the measure of an exterior angle on a triangle

nexus9112 [7]

Hi there!

There are many ways to find the measure of an exterior angle of a triangle. I'll explain the easiest method to you. As you can see in the image, a line is drawn extending a side of the triangle. In order to find the measure of the exterior angle, we can subtract the measure of the adjacent interior angle from 180 (a straight angle). Doing this subtraction will give us the measure of the exterior angle.

Hope this helps!! :)

6 0

3 years ago

Read 2 more answers

53% of 80<br> ? What is the percentage?

quester [9]

Answer:

42.4

Hope that this helps!

5 0

3 years ago

Read 2 more answers

Find the value of 2x+y when x=4 and y3

worty [1.4K]

Answer:

11

Step-by-step explanation:

Replace the values of x and y into the equation.

2(4) + 3 = 8 + 3

= 11

5 0

3 years ago

Read 2 more answers

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0

3 years ago