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2 years ago

Drag an expression or phrase to each box to complete the proof.

Mathematics

2 answers:

adelina 88 [10]2 years ago

5 0

Answer:

1st blank: ∠4≅∠1 and ∠3≅∠2

2nd blank: Angle-Angle similarity postulate

3rd blank: Substitution property of equality.

Step-by-step explanation:

Consider the provided figure.

Statement Reason

ΔACE, BD║AE Given

∠4≅∠1 and ∠3≅∠2 Corresponding angle postulate

ΔACE $\sim$ ΔBCD Angle-Angle similarity postulate

$\frac{CA}{CB}=\frac{CE}{CD}$ Definition of similar triangles

CA=CB+BA and CE=CD+DE Segment addition postulate

$\frac{CB+BA}{CB}=\frac{CD+DE}{CD}$ Substitution property of equality.

$\frac{CB}{CB}+\frac{BA}{CB}=\frac{CD}{CD}+\frac{DE}{CD}$ Addition of fractions.

$1+\frac{BA}{CB}=1+\frac{DE}{CD}$ Simplification of fractions

$\frac{BA}{CB}=\frac{DE}{CD}$ Subtraction property of equality.

QveST [7]2 years ago

3 0

<4 ≅ <1, 3 <≅ 2 Corresponding angles postulate

ΔACE ≈ ΔBCD Angle angle similarity postulate

(CB + BA)/CB = (CD + DE)/CD Substitution property of equality

You might be interested in

Simplify please! sqrt(8x^(12) y^(7))

HACTEHA [7]

$\sqrt{8 x^{12} y^{7} } = \sqrt{8} \sqrt{x^{12}} \sqrt{ y^{6}*y } =\sqrt{4} \sqrt{2} \sqrt{(x^{6} )^{2} } \sqrt{ (y^{3})^{2}} \sqrt{y}=2 \sqrt{2} x^{6} y^{3} \sqrt{y} 
$

I hope it helps you :)

8 0

2 years ago

Confused! Need help please!! Will mark brainliest!

ziro4ka [17]

Yo sup??

This question can be solved by applying the properties of similar triangles

the triangle with sides 5x and 20 is similar to the triangle with sides 45 and 35

The similarity property used here is called AAA ie angle angle angle property as all the three angles of the 2 triangles are equal.

therefore we can say

5x/45=20/36

x=5 units

Hope this helps

7 0

2 years ago

Find the perimeter: A. 2000xy + 200x B. 2200xy C. 1000xy + 200x D. 2200x

lapo4ka [179]

Should be A. Can’t combine X and Xy.

3 0

2 years ago

1) Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let x be a r

VMariaS [17]

Answer:

a) 0.9706 = 97.06% probability that x is less than 60.

b) 0.9987 = 99.87% probability that x is greater than 16.

c) 0.9693 = 96.93% probability that x is between 16 and 60.

d) 0.0294 = 2.94% probability that x is more than 60.

a) 0.0668 = 6.68% probability that the thickness is less than 3.0 mm.

b) 0.0062 = 0.62% probability that the thickness is more than 7.0 mm

c) 0.9270 = 92.70% probability that the thickness is between 3.0 mm and 7.0 mm.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

We have that $\mu = 43, \sigma = 9$

a) x is less than 60

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 43}{9}$

$Z = 1.89$

$Z = 1.89$ has a pvalue of 0.9706

0.9706 = 97.06% probability that x is less than 60.

b) x is greater than 16

This is 1 subtracted by the pvalue of Z when X = 16.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16 - 43}{9}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0013.

1 - 0.0013 = 0.9987

0.9987 = 99.87% probability that x is greater than 16.

c) x is between 16 and 60

This is the pvalue of Z when X = 60 subtracted by the pvalue of Z when X = 16.

From a), Z when X = 60 has a pvalue of 0.9706.

From b), Z when X = 16 has a pvalue of 0.0013

0.9706 - 0.0013 = 0.9693

0.9693 = 96.93% probability that x is between 16 and 60.

d) x is more than 60

This is 1 subtracted by the pvalue of Z when X = 60.

From a), Z when X = 60 has a pvalue of 0.9706.

1 - 0.9706 = 0.0294

0.0294 = 2.94% probability that x is more than 60.

Now $\mu = 4.5, \sigma = 1$

a) the thickness is less than 3.0 mm

This is the pvalue of Z when X = 3.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3 - 4.5}{1}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.0668 = 6.68% probability that the thickness is less than 3.0 mm.

b) the thickness is more than 7.0 mm

This is 1 subtracted by the pvalue of Z when X = 7.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7 - 4.5}{1}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938.

1 - 0.9938 = 0.0062

0.0062 = 0.62% probability that the thickness is more than 7.0 mm

c) the thickness is between 3.0 mm and 7.0 mm.

This is the pvalue of Z when X = 7 subtracted by the pvalue of Z when X = 3.

From b), Z when X = 7 has a pvalue of 0.9938.

From a), Z when X = 3 has a pvalue of 0.0668

0.9938 - 0.0668 = 0.9270

0.9270 = 92.70% probability that the thickness is between 3.0 mm and 7.0 mm.

8 0

2 years ago

Ricardo can type 39 words per minute. Estimate the number of words he can type in 18 minutes.

Vinvika [58]

Hello!

In order to find how many words Ricardo can type in 18 minutes, you'll need to subtract 39 by 18. You'll need to multiply because Ricardo is typing 39 words per minute for 18 minutes so you'll need to multiply to get the number of words he types for 18 minutes. Once you've multiplied 39 with 18, you should have a word count for 702 words for 18 minutes.

So Ricardo can type 702 words per minute.

I hope this helps!

7 0

2 years ago

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