All categories

3 years ago

PLEASE HELP!!! ILL GIVE BRAINLIEST !!

Mathematics

2 answers:

kow [346]3 years ago

8 0

Answer:

C. Alternate Interior angles.

Step-by-step explanation:

Alternate interior angles are inside parallel lines, and on opposite sides of the transversal line that crosses the parallel lines.

It's not A, because corresponding angles are in the same position on parallel lines. Angles 2 & 6 are corresponding angles. It's not B, because the two are not next to each other, forming a straight line. 6&7 are straight angles, and 5&6 are straight angles.

hoa [83]3 years ago

6 0

Answer:

look at my explanation..

Step-by-step explanation:

a. 1 and 3

b. 2 and 7

c. 2 and 8

You might be interested in

You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.

postnew [5]

This problem seems to be wrong because no minimum point was found and no point of landing exists

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

Derivatives

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

The function provided in the question is

$h(x)=2x^2-12x+20$

Let's find the first derivative

$h'(x)=4x-12$

solving h'=0:

$4x-12=0$

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

To find when will the ball land, we set h=0

$2x^2-12x+20=0$

Simplifying by 2

$x^2-6x+10=0$

Completing squares

$x^2-6x+9+10-9=0$

Factoring and rearranging

$(x-3)^2=-1$

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0

3 years ago

What is the sum of the polynomials?

lubasha [3.4K]

9x^3 - 8x^2 is the answer. to solve this you must distribute the sign across each parenthetical and then add the like terms (all cubed x variables added to cubed variables and all squared x variables added to squared variable)

6 0

3 years ago

Read 2 more answers

How do I write 8 more than k

lys-0071 [83]

Step-by-step explanation:

<h3>To write 8 more than follow this:-</h3><h3> k+8</h3>

6 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$