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4 years ago

15

Which best compares the pair of intersecting rays with the angle

1 answer:

madam [21]4 years ago

5 0

Answer:

C)The rays extend infinitely, and the angle is made by rays which have a common endpoint

Step-by-step explanation:

A ray is a line segment having one end point while extending in the other/opposite direction

An angle is made by two lines(rays) with common end point (vertex)

So when an angle is formed by the pair of intersecting rays the following options are ruled out

A) rays cannot have two end points

B) As rays extends infinitely in one direction, so having number of points is wrong

D)angles do not have lines

the following is best explanation is C

The rays extend infinitely, and the angle is made by rays which have a common endpoint !

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Please help solve these proofs asap!!!

timama [110]

Answer:

Proofs contained within the explanation.

Step-by-step explanation:

These induction proofs will consist of a base case, assumption of the equation holding for a certain unknown natural number, and then proving it is true for the next natural number.

a)

Proof

Base case:

We want to shown the given equation is true for n=1:

The first term on left is 2 so when n=1 the sum of the left is 2.

Now what do we get for the right when n=1:

$\frac{1}{2}(1)(3(1)+1)$

$\frac{1}{2}(3+1)$

$\frac{1}{2}(4)$

$2$

So the equation holds for n=1 since this leads to the true equation 2=2:

We are going to assume the following equation holds for some integer k greater than or equal to 1:

$2+5+8+\cdots+(3k-1)=\frac{1}{2}k(3k+1)$

Given this assumption we want to show the following:

$2+5+8+\cdots+(3(k+1)-1)=\frac{1}{2}(k+1)(3(k+1)+1)$

Let's start with the left hand side:

$2+5+8+\cdots+(3(k+1)-1)$

$2+5+8+\cdots+(3k-1)+(3(k+1)-1)$

The first k terms we know have a sum of .5k(3k+1) by our assumption.

$\frac{1}{2}k(3k+1)+(3(k+1)-1)$

Distribute for the second term:

$\frac{1}{2}k(3k+1)+(3k+3-1)$

Combine terms in second term:

$\frac{1}{2}k(3k+1)+(3k+2)$

Factor out a half from both terms:

$\frac{1}{2}[k(3k+1)+2(3k+2]$

Distribute for both first and second term in the [ ].

$\frac{1}{2}[3k^2+k+6k+4]$

Combine like terms in the [ ].

$\frac{1}{2}[3k^2+7k+4$

The thing inside the [ ] is called a quadratic expression. It has a coefficient of 3 so we need to find two numbers that multiply to be ac (3*4) and add up to be b (7).

Those numbers would be 3 and 4 since

3(4)=12 and 3+4=7.

So we are going to factor by grouping now after substituting 7k for 3k+4k:

$\frac{1}{2}[3k^2+3k+4k+4]$

$\frac{1}{2}[3k(k+1)+4(k+1)]$

$\frac{1}{2}[(k+1)(3k+4)]$

$\frac{1}{2}(k+1)(3k+4)$

$\frac{1}{2}(k+1)(3(k+1)+1)$ .

Therefore for all integers n equal or greater than 1 the following equation holds:

$2+5+8+\cdots+(3n-1)=\frac{1}{2}n(3n+1)$

//

b)

Proof:

Base case: When n=1, the left hand side is 1.

The right hand at n=1 gives us:

$\frac{1}{4}(5^1-1)$

$\frac{1}{4}(5-1)$

$\frac{1}{4}(4)$

$1$

So both sides are 1 for n=1, therefore the equation holds for the base case, n=1.

We want to assume the following equation holds for some natural k:

$1+5+5^2+\cdots+5^{k-1}=\frac{1}{4}(5^k-1)$ .

We are going to use this assumption to show the following:

$1+5+5^2+\cdots+5^{(k+1)-1}=\frac{1}{4}(5^{k+1}-1)$

Let's start with the left side:

$1+5+5^2+\cdots+5^{(k+1)-1}$

$1+5+5^2+\cdots+5^{k-1}+5^{(k+1)-1}$

We know the sum of the first k terms is 1/4(5^k-1) given by our assumption:

$\frac{1}{4}(5^k-1)+5^{(k+1)-1}$

$\frac{1}{4}(5^k-1)+5^k$

Factor out the 1/4 from both of the two terms:

$\frac{1}{4}[(5^k-1)+4(5^k)]$

$\frac{1}{4}[5^k-1+4\cdot5^k]$

Combine the like terms inside the [ ]:

$\frac{1}{4}(5 \cdot 5^k-1)$

Apply law of exponents:

$\frac{1}{4}(5^{k+1}-1)$

Therefore the following equation holds for all natural n:

$1+5+5^2+\cdots+5^{n-1}=\frac{1}{4}(5^n-1)$ .

//

5 0

3 years ago