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1 year ago

Please help me! Which congruence statement does not necessarily describe the triangle shown if DEF equals FGH?

Mathematics

1 answer:

andre [41]1 year ago

7 0

ΔEDF ≅ ΔFGH in this congruence statement does not necessarily describe the triangle shown if DEF equals FGH, so, ΔEDF ≅ ΔFGH is correct answer.

<h3>What is a Congruent triangles :</h3>

Two triangles are said to be congruent if all three corresponding sides are equal and all the three corresponding angles are equal in measure. These triangles can be slides, rotated, flipped and turned to be looked identical. If repositioned, they coincide with each other.

Based on the given conditions,

Since,

ΔDEF ≅ ΔFGH

Then,

∠FED ≅ ∠HGF

∠EFD ≅ ∠GHF

∠EDF ≅ ∠GFH

Congruent triangles corresponding to equal angles,

Therefore,

ΔEDF ≅ ΔFGH in this congruence statement does not necessarily describe the triangle shown if DEF equals FGH.

To learn more about information visit Congruence triangle :

brainly.com/question/12248294

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Tommy has 6 more than 3 times the amount of money in his bank account than Judy. Judy has $7500 more than a number in her accoun

Ad libitum [116K]

Answer:

Amount of money Judy has in her account is $\$(7500+n)$ .

Amount of money Tommy has in his account is $\$(22506+3n)$

Step-by-step explanation:

To find : Amount of money in Judy account and amount of money in Tommy account:

Solution:

Given:

Judy has $7500 more than a number in her account.

Let the number be 'n'.

So we can say that;

Amount of money in Judy account is equal to 7500 plus number.

framing in equation form we get;

Amount of money in Judy account = $\$(7500+n)$

Hence Amount of money Judy has in her account is $\$(7500+n)$ .

Now Given:

Tommy has 6 more than 3 times the amount of money in his bank account than Judy.

So we can say that;

Amount of money in Tommy account is equal to 3 multiplied by Amount of money in Judy account plus 6.

framing in equation form we get;

Amount of money in Tommy account = $3(7500+n)+6 =22500+3n+6=\$(22506+3n)$

Hence Amount of money Tommy has in his account is $\$(22506+3n)$ .

8 0

3 years ago

HELP!! What is the approximate measure of angle x in the triangle shown?

Alex Ar [27]

<h3>Answer: D) 130.5 degrees</h3>

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Work Shown:

$c^2 = a^2 + b^2 - 2*a*b*\cos(C)\\\\10^2 = 5^2 + 6^2 - 2*5*6*\cos(x)\\\\100 = 25 + 36 - 60*\cos(x)\\\\100 = 61 - 60*\cos(x)\\\\100 - 61 = - 60*\cos(x)\\\\39 = - 60*\cos(x)\\\\\cos(x) = \frac{39}{-60}\\\\\cos(x) = -0.65\\\\x = \arccos(-0.65)\\\\x \approx 130.5416\\\\x \approx 130.5\\\\$

Note: I used the law of cosines. Make sure your calculator is in degree mode.

6 0

3 years ago

Living with parents: The Pew Research Center reported that 36% of American Millennials (adults ages 18-31) stll live at home wit

Lera25 [3.4K]

Answer:

$\mathbf{The \ sample \ comes \ from \ a \ population \ of \ Millennial \ students }$ $\mathbf{at \ their \ campus\ where \ 3 6\%\ still\ live\ at\ home\ with\ their \ parents}$

Step-by-step explanation:

From the given information.

The proportion of American Millennials still with their parents = 0.36

The sample size = 300

Sample proportion = 0.43

Level of significance = 0.006

P-value = 0.006

Null hypothesis:

$\mathbf{H_o: Of \ Millennial \ students \ at \ their \ campus, \ 36\% \ }$ $\mathbf{ live \ at \ some \ with \ their \ parents . }$

$\mathbf{H_a: More \ than \ 36\% \ of \ Millennial \ students \ at \ their \ campus \ live \ at \ home }$ $\mathbf{ \ with \ their \ parents}$

The required task is to determine the assumption about the sample that underlies the hypothesis test from the given options.

$\mathbf{The \ sample \ comes \ from \ a \ population \ of \ Millennial \ students }$ $\mathbf{at \ their \ campus\ where \ 3 6\%\ still\ live\ at\ home\ with\ their \ parents}$

This is because the student wants to check if the null hypothesis ( which states that of Millennial students at their campus, 36% live at home with their parents) is correct or not.

7 0

3 years ago

The distance between two schools,A and B is 2 km.A market is situated a third of the distance from A to B.How far is the market

bekas [8.4K]

Answer:

$\frac{4}{3}$

Step-by-step explanation:

Find distance between B and market: |B-M|=|M-B|

Distance between A and B: $|A-B| = 2$ , call this equation (1)

Distance between A and market $|A-M| = \frac{1}{3} |A-B|=\frac{2}{3}$ , call this equation (2)

Solve (1) - (2)

$|A-B-A+M| = 2-\frac{2}{3} \\|M-B|=|B-M|=\frac{4}{3}$

3 0

3 years ago

Simplify each expression. Assume that all variables are positive.

kozerog [31]

Q1. The answer is $\frac{8x^{3}y^{6} }{27}$

$( \frac{16 x^{5} y^{10}}{81x y^{2} } )^{ \frac{3}{4} }= ( \frac{16}{81}* \frac{ x^{5} }{x}* \frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} } \\ \\ 
 \frac{ x^{a} }{ x^{b} }= x^{a-b} \\ \\ 
( \frac{16}{81}* \frac{ x^{5} }{x}*\frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} }}=( \frac{16}{81 }* x^{5-1}* y^{10-2})^{ \frac{3}{4} }=( \frac{16}{81 }* x^{4}* y^{8})^{ \frac{3}{4} }= \\ \\ = (\frac{16}{18} )^{ \frac{3}{4} }*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=$
$\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\ \\ 
 (x^{a} )^{b} = x^{a*b} \\ \\ 
\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } = \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} =$
$= \frac{8x^{3}y^{6} }{27}$

Q2. The answer is 1/16

$(-64) ^ \frac{-2}{3} =(-1* 2^{6} ) ^ \frac{-2}{3}=(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} \\\\x^{-a} = \frac{1}{ x^{a} } \\\\(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} = \frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{(2^{6})^ \frac{2}{3}} \\ \\ (x^{a} )^{b}=x^{a*b} \\\\x^{ \frac{a}{b} = \sqrt[b]{ x^{a} } } \\ \\ 
$
$\frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{2^{6*\frac{2}{3}}} = \frac{1}{ \sqrt[3]{(-1)^{2} } } * \frac{1}{ 2^{4} } = \frac{1}{ \sqrt[3]{1} } * \frac{1}{16} = \frac{1}{1} * \frac{1}{16}= \frac{1}{16}$

Q3. The answer is $a^{ \frac{7}{6} }$

$a^{ \frac{2}{3} } * a^{ \frac{1}{2} } \\ \\ 
 x^{a}* x^{b} =x^{a+b} \\ \\ 
a^{ \frac{2}{3} } * a^{ \frac{1}{2} }= a^{ \frac{2}{3} + \frac{1}{2} } =a^{ \frac{2*2}{3*2} + \frac{1*3}{2*3} }=a^{ \frac{4}{6} + \frac{3}{6} }=a^{ \frac{4+3}{6} }=a^{ \frac{7}{6} }$

7 0

3 years ago