Ask question

Ask question

All categories

2 years ago

15

Quadrilateral A’B’C’D’ is a dilation of quadrilateral ABCD about point P with a scale factor of 2 1/2 .

2 answers:

mina [271]2 years ago

8 0

Answer:

A. Enlargement

Step-by-step explanation:

We are given that,

Quadrilateral ABCD is dilated by a scale factor of $2\frac{1}{2}=\frac{5}{2}$ .

Now, we know that,

<h3>Dilation changes the size of the figure by the scale factor 'k'.</h3>

<em>If 'k' > 1, then the dilated figure is enlarged in size.</em>

<em>If 0 < 'k' < 1, then the dilated figure is reduced in size.</em>

Since, the given scale factor is $\frac{5}{2}=2.5>1$ .

<h3>So, the quadrilateral ABCD is enlarged in size.</h3>

Hence, option A is correct.

kenny6666 [7]2 years ago

7 0

A it says can you tell me why??

You might be interested in

The sum of two polynomials is –yz2 – 3z2 – 4y + 4. If one of the polynomials is y – 4yz2 – 3, what is the other polynomial? –2yz

marishachu [46]

The first polynomial is $y -4yz^2-3$ and let the second polynomial be P(y,z).

If the sum of two polynomials is $-yz^2-3z^2-4y+4$ , then $y-4yz^2-3+P(y,z)=-yz^2-3z^2-4y+4$ .

Take the first polynomial from the left side to the right side:

$P(y,z)=-yz^2-3z^2-4y+4 -(y-4yz^2-3),\\ P(y,z)=-yz^2-3z^2-4y+4 -y+4yz^2+3,\\ P(y,z)=3yz^2-3z^2-5y+7$ .

Answer: the second polynomial is $3yz^2-3z^2-5y+7$ .

3 0

3 years ago

Read 2 more answers

15 workers working 8hrs a day can complete a piece of work in 20 days. In how many days will 10 workers do it,working 6 hrs a da

mel-nik [20]

Answer:

40 days

Step-by-step explanation:

w1*h1*d1=w2*h2*d

15*8*20=10*6*d

2400=60d

d=2400/60

d=40

6 0

2 years ago

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago

The average rate of change of a function f(x) = x2 between x = 3 and x = 9 is

masya89 [10]

Answer:

12.

Step-by-step explanation:

Average rate of change = (value of the function when x = 9 - value when x=3)

/

(9 - 3)

= f(9) - f(3) / (9-3)

= (9^2 - 3^2) / (9-3)

= 72 / 6

= 12.

6 0

2 years ago

What is 2(6+7)= 3(5+2)

Komok [63]

Solution:

2(6+7)= 3(5+2)

Step ; 1 { Solving first Term i.e LHS }

LHS : 2 ( 6 + 7)

☞ 2 ( 13)

☞ 2 × 13

☞ 26

Now, RHS ; 3 ( 5 + 2)

☞ 3 ( 7)

☞ 21

From Equation 1 and 2, We can LHS and RHS are not Equal.

Therefore, 26≠ 21.

7 0

3 years ago