All categories

1 year ago

ΔABC is dilated using a scale factor of 12 to produce ΔA'B'C'. Select all of the statements that apply to the transformation.

Mathematics

1 answer:

GalinKa [24]1 year ago

3 0

Answer:

Options (3) and (6)

Step-by-step explanation:

ΔABC is a dilated using a scale factor of $\frac{1}{2}$ to produce image triangle ΔA'B'C'.

Since, dilation is a rigid transformation,

Angles of both the triangles will be unchanged or congruent.

m∠A = m∠A' and m∠B = m∠B'

Since, sides of ΔA'B'C' = $\frac{1}{2}$ of the sides of ΔABC

Area of ΔA'B'C' = $\frac{1}{2}(\text{Area of triangle ABC})$

Area of ΔABC > Area of ΔA'B'C'

Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.

ΔABC ~ ΔA'B'C'

Therefore, Option (3) and Option (6) are the correct options.

You might be interested in

A home improvement store sold wind chimes for $30 each. A customer signed up for a free membership card and received a 5% discou

dsp73

the final price of the wind chime is $29.925 .

<u>Step-by-step explanation:</u>

Here we have , A home improvement store sold wind chimes for $30 each. A customer signed up for a free membership card and received a 5% discount off the price. Sales tax of 5% was applied after the discount. We need to find What was the final price of the wind chime . Let's find out:

Initially we have , 5% discount in $30 i.e.

⇒ (30(5))/100

⇒ (150)/100

⇒ $1.5 , So now price becomes $30 - $1.5 = $28.5 . At this amount there's 5% tax i.e.

⇒ (28.5(5))/100

⇒ (142.5)/100

⇒ $1.425

So now price becomes $28.5 + $1.425 = $29.925 . Therefore , the final price of the wind chime is $29.925 .

3 0

2 years ago

Isosceles △ABC (AC=BC) is inscribed in the circle k(O). Prove that the tangent to the circle at point C is parallel to AB .

timofeeve [1]

Explanation:

Let M be the midpoint of AB. Then CM is the perpendicular bisector of AB. As such, center O is on CM, and OC is a radius (and CM). The tangent is perpendicular to that radius (and CM), so is parallel to AB, which is also perpendicular to CM.

If you need to go any further, you can show that triangles CMA and CMB are congruent, so (linear) angles CMA and CMB are congruent, hence both 90°.

5 0

2 years ago

A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain

svetlana [45]

Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

$\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55$

The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

$LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8$

The probability of having between 4 and 8 customers choosing the oversize version is:

$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

$P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328$

7 0

2 years ago

A cone-shaped pile of gravel has a diameter of 30 m and a height of 9.1 m.

Arte-miy333 [17]

D/2=r
vcone=(1/3)hpir^2
given
d=30
h=9.1
d/2=30/2=15=r

v=(1/3)9.1pi15^2
v=(9.1/3)pi225
v=682.5pi
use 3.141592 to aprox pi
v=2144.13654
the closese is the first one
answer would be 2120 m³

6 0

2 years ago

Read 2 more answers

381.10-214.43 equals

dmitriy555 [2]

Answer:

166.67

Step-by-step explanation:

3 0

2 years ago