90/ [10(3squared

Nina [5.8K]

X= 70 degrees
Y= 70 degrees

Understand that every triangle has three angles and they add up to 180 degrees.

If I split this triangle in half the total degrees of each individual piece will be 90 degrees. A split in the isosceles triangle will also cause the 40 degrees to halved (thus, how I got 20 degrees in our 90 triangle).

Since we are dealing with an isosceles triangles two of the sides will be equal (hence, the dashes on the triangles sides). Therefore, x and y will also be equal.

Now if our 40 degreed angle is now 20 degrees, we have an unknown angle and the triangle in total now adds up to 90 degrees we can set up an equation.

20 + y = 90

Y = 70

Since X and Y are equal, X will also be 70.

If we return to to the isosceles triangle before it was split (use your photo for reference) and we add 40 +70 + 70 we will get 180 degrees. Which is the standard total of degrees for any triangle that is not a 90 degreed triangle.

I hope this helps. Feel free to ask questions.

Below I uploaded my work.

8 0

1 year ago

When given Three points on a coordinate plane, you may be able to find a specific Quadratic equation that intersects all given p

ohaa [14]

Answer:

TRUE

Step-by-step explanation:

A quadratic equation can be found that will go through any three distinct points that ...

satisfy the requirements for a function
are not on the same line

_____

The key word here is "may." You will not be able to find a quadratic intersecting the three points if they do not meet both requirements above.

8 0

3 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

Use the points in the diagram to name the figure.

Viefleur [7K]

D) Line JK

<----->

JK

Hope it helps :)

3 0

1 year ago

What is the measure of Angle B A C?

Simora [160]

It's 180-50-50=180-100=80°

8 0

3 years ago

90/ [10(3squared - 4)