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3 years ago

Which equation can be used tº find B in the triangle below?

Mathematics

1 answer:

Anton [14]3 years ago

6 0

Answer:

tanB=8/6

Step-by-step explanation:

Read more about number lines at:

brainly.com/question/4727909

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4 0

2 years ago

In an instant lottery, your chances of winning are 0.1. if you play the lottery five times and outcomes are independent, the pro

lesya692 [45]

The probability that you lose all five times will be 0.59.

<h3>How to calculate the probability?</h3>

From the information given, the chances of winning are 0.1. Therefore, the chance of losing will be:

= 1 - 0.1 = 0.9

Therefore, the probability that you lose all five times will be:

= 0.9 × 0.9 × 0.9 × 0.9 × 0.9

= 0.59

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6 0

2 years ago

Apollo 8 orbited the moon in a circular orbit. Its average altitude was 185 km above the moon's surface. Create an equation to m

Marta_Voda [28]

You probably haven't learnt this but the general equations for circle with centre at the origin is x^2+y^2=R^2 where R is the radius of the circle so R^2=(1737+185)^2.I am lazy to press my calculator so type it in yourself.

8 0

2 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

Anyone know the answer?

aleksklad [387]

I want to say it’s the second one

3 0

2 years ago