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goldfiish [28.3K]

3 years ago

11

True or false the pre-image and image of a dilation are always similar.

1 answer:

Dmitry [639]3 years ago

7 0

False: although the size increases to its ratio, it is not considered similar

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Consider the initial value problem:

Inessa [10]

Answer:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

Step-by-step explanation:

The given initial value problem is;

$y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2$

Let

$u(t)=y(t)----(2)\\v(t)=y'(t)----(3)$

Differentiating both sides of equation (1) with respect to $t$ , we obtain:

$u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)$

Differentiating both sides of equation (2) with respect to $t$ gives:

$v'(t)=y''(t)----(6)$

From equation (1),

$y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)$

Putting t=0 into equation (2) yields

$u(0)=y(0)\\\implies u(0)=-5$

Also putting t=0 into equation (3)

$u'(0)=y'(0)\\u'(0)=2$

The system of first order equations is:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

3 0

3 years ago

To ensure that the horizontal segments are parallel, x must equal

andrezito [222]

Answer:

53

Step-by-step explanation:

Because its 53.

6 0

3 years ago

i sell a jacket i bought for 200 dollars at a 20% increase. how much money will i make off of selling it

Ann [662]

Answer:

$40

Step-by-step explanation:

4 0

3 years ago

If your cell phone bill was 67 Point $82 and there is a 7.5% late fee how much will your bill be with the late fee included

Ivenika [448]

$\frac{7.5}{100} = \frac{x}{82} \\ x = \frac{82 \times 7.5}{100} = \frac{615}{100 } = 6.15$
$82 + 6.15 = 88.15$
the answer is $88.15

7 0

3 years ago

Find the value of tan theta if sin theta = 12/13 and theta is in quadrant 2

8090 [49]

Answer:

tanΘ = - $\frac{12}{5}$

Step-by-step explanation:

Using the trigonometric identities

• sin²x + cos²x = 1, hence

cosx = ± √(1 - sin²x )

• tanx = $\frac{sinx}{cosx}$

given sinΘ = $\frac{12}{13}$ , then

cosΘ = ± $\sqrt{1-(12/13)^2}$

Since Θ is in the second quadrant where cosΘ < 0, then

cosΘ = - $\sqrt{1-\frac{144}{169} }$

= - $\sqrt{\frac{25}{169} }$ = - $\frac{5}{13}$

tanΘ = $\frac{\frac{12}{13} }{\frac{-5}{13} }$

= $\frac{12}{13}$ × - $\frac{13}{5}$ = - $\frac{12}{5}$

5 0

3 years ago