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4 years ago

Card 5

Mathematics

1 answer:

lord [1]4 years ago

8 0

Scale Factor to go from F to G = 3/4

Scale Factor to go from G to F = 4/3

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Explanation:

Note the sides labeled 4 and 3 are both opposite the angles that have two tickmarks. Therefore, these two sides correspond to one another.

Because of the similar triangles, the corresponding sides are in proportion. The scale factor is 3/4 meaning the smaller triangle has side lengths 3/4 as large compared to the larger triangle.

Put another way, we have a 25% reduction in the side lengths because 3/4 = 75% and 100% - 75% = 25%

So far, all of this applies when figure F is the preimage and G is the image. If we reverse things, then apply the reciprocal to 3/4 to get 4/3. Now G is the preimage and we go from smaller to larger.

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Side notes:

If the scale factor is smaller than 1, but still positive, then the image is smaller than the preimage. We have an image reduction.
If the scale factor is larger than 1, then the image is larger than the preimage. We have an image enlargement.

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The price of a car was bought for $20,000 and has depressive 15% yearly find the price of the car in 6 years

a_sh-v [17]

Answer:

7542.99

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3 years ago

Can somebody help me with this What is the value of P

leonid [27]

Answer:

p=7

∠BRG=49

Step-by-step explanation:

∠BRG and ∠YRG are on a straight line together. Therefore, they are supplementary and add to 180 degrees.

∠BRG+∠YRG=180

We know that ∠BRG is 7p and ∠YRG is 131 degrees. Substitute 7p in for ∠BRG and 131 for ∠YRG.

7p+131=180

We want to find out what p is. In order to do that, we have to get p by itself. First, subtract 131 from both sides.

7p+131-131=180-131

7p=49

Next, divide both sides by 7.

7p/7=49/7

p=49/7

p=7

Now, we have to find what ∠BRG is.

∠BRG=7p

We know that p=7, so substitute p in for 7

∠BRG=7*7

∠BRG=49

So, p=7 and ∠BRG=49

6 0

3 years ago

Read 2 more answers

Solve. 3/2x – 4 = 16

Luda [366]

$\dfrac{3}{2x}-4=16\ \ \ |add\ 4\ to\ both\ sides\\\\\dfrac{3}{2x}=20\\\\\dfrac{3}{2x}=\dfrac{20}{1}\ \ \ |cross\ multiply\\\\20\cdot2x=3\cdot1\\\\40x=3\ \ \ |divide\ both\ sides\ by\ 40\\\\\boxed{x=\frac{3}{40}}$

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3 years ago

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Simplify. 4 to the power of 3 + 4x5 •80 •84 •240 •340

djyliett [7]

4^3 + 4*5

4^3 = 64

64 + 4*5

4*5 = 20

64 + 20 = 84

84 is the simplest answer.

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4 years ago

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

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3 years ago