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2 years ago

Are these figures are similar? If so, what is the scale factor?

Mathematics

1 answer:

tester [92]2 years ago

8 0

Answer:

Yes, the scale factor is 0.6 repeating, or 0.7.

Step-by-step explanation:

You find the scale factor by dividing the smaller sides by the corresponding larger ones.

12/18 = 0.6 repeating

14/21 = 0.6 repeating

16/24 = 0.6 repeating

These triangles could be similar by SSS or SAS or ASA I think since congruent angles are shown in both triangles as well.

You might be interested in

Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

5 0

3 years ago

A student constructs a regular square pyramid with a base that has sides of 13.2 cm and a slant height of 6 cm. Find the surface

Komok [63]

Answer:

It is 409.72.

Step-by-step explanation:

6 0

3 years ago

Ricky race $640 in a charity walk last year this year he raised 15% more than he raised last year how much money did Ricky raise

Mashutka [201]

He raised $736.

One way is to multiply 640 x .15 = 96
Add 96 + 640 = 736

8 0

3 years ago

How do I get the answer to 2/5 times 3

NARA [144]

Answer:

6/5

Step-by-step explanation:

2/5*3=2/5*3/1=6/5

3 0

3 years ago

Read 2 more answers

Suppose that only 25% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Juliette [100K]

Answer:

$X \sim Binom(n=15, p=0.25)$

For this case we can use the probability mass function and we got:

$P(X= 5) = (15C5) (0.25)^{5} (1-0.25)^{15-5}= 0.165$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=15, p=0.25)$

For this case we can use the probability mass function and we got:

$P(X= 5) = (15C5) (0.25)^{5} (1-0.25)^{15-5}= 0.165$

5 0

3 years ago