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Kisachek [45]
3 years ago
7

Perform the indicated operation. 2 1/6 – 5 2/3

Mathematics
2 answers:
REY [17]3 years ago
6 0
<span>2 1/6 – 5 2/3
=</span><span>2 1/6 – 5 4/6
= - 3 3/6
=  -3 1/2

answer
</span><span>C. -3 1/2</span>
saul85 [17]3 years ago
3 0

Answer:

Option C - 2\frac{1}{6}-5\frac{2}{3}=-3\frac{1}{2}

Step-by-step explanation:

Given : Expression 2\frac{1}{6}-5\frac{2}{3}

To find : Perform the indicated operation in the expression ?

Solution :

Expression 2\frac{1}{6}-5\frac{2}{3}

Write mixed fraction into fraction,

2\frac{1}{6}-5\frac{2}{3}=\frac{13}{6}-\frac{17}{3}

Taking Least common denominator,

2\frac{1}{6}-5\frac{2}{3}=\frac{13-34}{6}

2\frac{1}{6}-5\frac{2}{3}=-\frac{21}{6}

2\frac{1}{6}-5\frac{2}{3}=-\frac{7}{2}

Write improper fraction into mixed fraction,

2\frac{1}{6}-5\frac{2}{3}=-3\frac{1}{2}

Therefore, Option C is correct.

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Which reason justifies step 2 in the proof
faust18 [17]

Answer:

Step-by-step explanation:

Given: ∠N≅∠S, line l bisects TR at Q.

To prove: ΔNQT≅ΔSQR

Proof:

From  ΔNQT and ΔSQR

It is given that:

∠N≅∠S (Given)

∠NQT≅∠SQR(Vertical opposite angles)

and TQ≅QR ( Definition of segment bisector)

Thus, by AAS rule,

ΔNQT≅ΔSQR

Hence proved.

Statement                                                 Reason

1. ∠N≅∠S                                                    given

2. ∠NQT≅∠SQR                            Vertical angles are congruent

3.  line l bisects TR at Q.                            given

4. TQ≅QR                                      Definition of segment bisector

5. ΔNQT≅ΔSQR                           AAS theorem

Hence proved.

Thus, option D is correct.

3 0
3 years ago
A small school has 42 boys. If the ratio of boys to girls is 7 : 1, how many, students are?
babunello [35]
There are a total of 48 students. Being 42 boys as mentioned earlier and 6 girls total attending too.

Please vote my answer brainliest! Thanks.
8 0
3 years ago
Read 2 more answers
Evaluate the surface integral. s y ds, s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v
Juliette [100K]

Compute the surface element:

\mathrm dS=\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv

\vec r(u,v)=(u\cos v,u\sin v,v)\implies\begin{cases}\vec r_u=(\cos v,\sin v,0)\\\vec r_v=(-u\sin v,u\cos v,1)\end{cases}

\|\vec r_u\times\vec r_v\|=\sqrt{\sin^2v+(-\cos v)^2+u^2}=\sqrt{1+u^2}

So the integral is

\displaystyle\iint_Sy\,\mathrm dS=\int_0^\pi\int_0^6u\sin v\sqrt{1+u^2}\,\mathrm du\,\mathrm dv

=\displaystyle\left(\int_0^\pi\sin v\,\mathrm dv\right)\left(\int_0^6u\sqrt{1+u^2}\,\mathrm du\right)

=\dfrac23(37^{3/2}-1)

4 0
3 years ago
You borrow $2,000 for a period of 4 years you are charged simple interest at a rate of 3% how much will you repay at the end of
Aleks [24]
At the end of 4 years you will pay a extra 240 dollars 
8 0
3 years ago
Indicate by true or false whether the following equation is quadratic -3/4x^2 +2=0 true or false
yKpoI14uk [10]

\bf -\cfrac{3}{4}x^2+2=0\impliedby \textit{notice the }^2\textit{, is \underline{quadratic}}


bear in mind that it goes with the "degree" of the polynomial.

5 0
3 years ago
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