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3 years ago

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Two triangles are similar. The sides of the first triangle are 8,10 and 12. The smallest side of the second triangle is 24. Find

the perimeter of the second triangle.

1 answer:

RSB [31]3 years ago

5 0

So 8*3=24, so multiply all of the sides by 3,
the other 2 are 30 and 36

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Free_Kalibri [48]

Answer:

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Step-by-step explanation:

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2 years ago

David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF o

BlackZzzverrR [31]

Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :( $b^4,b^2,b^4$ )

$b^4$ = b × b × b × b

$b^2$ = b × b

$b^4$ =b × b × b × b

GCF of variables :( $b^4,b^2,b^4$ ) is $b^2$

3) GCF of $c^3$ and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.

4) $80b^4-32b^2c^3+48b^4c$

$=16b^2(5b^2-2c^3+3b^2c)$

David applied the distributive property.

7 0

2 years ago

Read 3 more answers

g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa

adoni [48]

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is $\sigma = 0.0133$

Step-by-step explanation:

From the question we are told that

The upper specification is $USL = 1.68 \ mm$

The lower specification is $LSL = 1.52 \ mm$

The sample mean is $\mu = 1.6 \ mm$

The standard deviation is $\sigma = 0.03 \ mm$

Generally the capability index in mathematically represented as

$Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]$

Now what min means is that the value of CPk is the minimum between the value is the bracket

substituting value given in the question

$Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]$

=> $Cpk = min[ 0.88 , 0.88 ]$

So

$Cpk = 0.88$

Now from the question we are asked to evaluated the value of standard deviation that will produce a capability index of 2

Now let assuming that

$\frac{\mu - LSL }{ 3 * \sigma } = 2$

So

$\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2$

=> $0.08 = 6 \sigma$

=> $\sigma = 0.0133$

So

$\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }$

=> $2$

Hence

$Cpk = min[ 2, 2 ]$

So

$Cpk = 2$

So $\sigma = 0.0133$ is the value of standard deviation required

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Answer:

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Step-by-step explanation:

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3 years ago

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That's for the equation I got from the picture

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