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3 years ago

Two angles in a triangle add up to 85 degrees what is the size of the third angle

1 answer:

6 0

All of the angles in a triangle should add up to 180 degrees so if you know 2 out of three you just do subtract 180 by the sum of the two angles.
180-85=95

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What is 7n + 3 = 3n + 27

icang [17]

7n + 3 = 3n + 27

7n -3n =27-3
4n =24
n= 6.

4 0

3 years ago

Find the value of c and AB if B is between A and C,AB=2r,AC=-5and BC=7

elena-14-01-66 [18.8K]

AB=2r

BC=7

AC=AB+BC

-5=2r+7

-12=2r

-6=r

AB=-12

6 0

4 years ago

9.6x10^85/3x10^63 in scientific notation

Novosadov [1.4K]

Answer:

3.27*10^22

Step-by-step explanation:

Given the expression 9.6x10^85/3x10^63, we are to write it on scientific notation as shown:

9.6x10^85/3x10^63

= (9.8/3) * (10^85/10^63)

= (9.8/3) * 10^{85-63}

= (9.8/3) *10^22

= 3.27 *10^22

Hence the expression in scientific notation is 3.27*10^22

6 0

3 years ago

Hey everybody I have a quick math question but idk what the answer is! It would mean a lot if someone helped me. PS the pic is t

mote1985 [20]

Answer:

it would be 2.33 I divided it I hope I'm right :)

8 0

3 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago