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Sav [38]
1 year ago
13

I need some help please out

Mathematics
1 answer:
IceJOKER [234]1 year ago
7 0

Question:

Solution:

Let the following equation:

\sqrt[]{12-x}=\text{ x}

this is equivalent to:

(\sqrt[]{12-x})^2=x^2

this is equivalent to:

12-x=x^2

this is equivalent to:

x^2+x-12=\text{ 0}

thus, we can conclude that

x= 3.

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klasskru [66]
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3 years ago
It is not possible to prove one pair of triangles congruent and then use their congruent corresponding parts to prove another pa
borishaifa [10]

Answer:

true

The wording does not quite mean anything,

but what I think was meant to ask is

"if we use some parts of two triangles to prove they are congruent,

can we then use that to prove that

a pair of corresponding parts not used before are congruent?"

The answer is

Yes, of course,

Corresponding Parts of Congruent Triangles are Congruent,

which teachers usually abbreviate as CPCTC.

For example, if we find that

side AB is congruent with side DE,

side BC is congruent with side EF, and

angle ABC is congruent with angle DEF,

we can prove that triangles ABC and DEF are congruent

by Side-Angle-Side (SAS) congruence.

We then, by CPCTC, can conclude that other pairs of corresponding parts are congruent:

side AB is congruent with side DE,

angle BCA is congruent with angle EFD, and

angle CAB is congruent with angle FDE.

It was possible (by CPCTC) to prove those last 3 congruence statements,

after proving the triangles congruent.

The expected answer is FALSE.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Last week, Thomas jogged a total of 35.209 miles on his treadmill. Two weeks ago, he jogged 38.074 miles.
Sonja [21]

Answer:

2.865

Step-by-step explanation:

pls rate five stars, thanks  :)

6 0
3 years ago
Help please will be marked as brainliest if correct and first person to respond
nydimaria [60]

Answer: Unlock code = 76

==========================================================

Explanation:

Use a factor tree to find the prime factorization of each value

22 = 2*11

30 = 2*3*5

Both values have '2' in common and this is the largest such value. Therefore, the GCF of 22 and 30 is 2.

So E = 2

----------------------------

Since 3 and 8 have no factors in common (other than 1), this means the LCM is 3*8 = 24

We can say that F = 24.

If the numbers had some larger number in common, then we'd have to do a bit more work to find the LCM.

----------------------------

As the hint suggests, we'll use the distributive property to show that

3*(7+10) = G + 30

3*7 + 3*10 = G + 30

21 + 30 = G + 30

21 = G .... subtract 30 from both sides; they cancel and go away

G = 21

----------------------------

Use the same idea for the other equation as well

5*(3 + H) = 15 + 45

5*3 + 5*H = 15 + 45

15 + 5H = 45

5H = 45 ............ subtract 15 from both sides; they go away

H = 45/5 ......... divide both sides by 5

H = 9

------------------------------

The unlock code is

E+F+G+H = 2+24+21+29 = 76

4 0
3 years ago
Read 2 more answers
Help me pls due today
balandron [24]

Answer:

i really dont know but i hope you get it right good luck

Step-by-step explanation:

6 0
2 years ago
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