Explore topic-wise InterviewSolutions in Current Affairs.

HOPE this is helpful

Please MARK it as BRAINLIST ANSWER

ANSWER:

a+(a-1)*d..............

20+20+30+30=100 is the ANSWER

ANSWER:

Option (a) will be the REQUIRED answer

Step-by-step explanation:

PLEASE MARK me as the Brainliest answer

Answer:

please follow me

Step-by-step explanation:

STEP

1

:

Equation at the end of step 1

(11x2 + 18x) + 7

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring 11x2+18x+7

The first term is, 11x2 its coefficient is 11 .

The middle term is, +18x its coefficient is 18 .

The last term, "the constant", is +7

Step-1 : Multiply the coefficient of the first term by the constant 11 • 7 = 77

Step-2 : Find two FACTORS of 77 whose sum equals the coefficient of the middle term, which is 18 .

-77 + -1 = -78

-11 + -7 = -18

-7 + -11 = -18

-1 + -77 = -78

1 + 77 = 78

7 + 11 = 18 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 7 and 11

11x2 + 7x + 11X + 7

Step-4 : Add up the first 2 TERMS, pulling out like factors :

x • (11x+7)

Add up the last 2 terms, pulling out common factors :

1 • (11x+7)

Step-5 : Add up the FOUR terms of step 4 :

(x+1) • (11x+7)

Which is the desired factorization

Final result :

(11x + 7) • (x + 1)

ANSWER:

THANKS for FREE points

thxxxxxxxxxxxxxxxxxxxxxxx

ANSWER:

Sorry

sorry

Step-by-step EXPLANATION:

I haven't have study about this TOPIC

plz plz plz FOLLOW me

ANSWER:

238.09

Step-by-step explanation:

238.09 is the FINAL answer

THANK me LATER

Answer:

1

Step-by-step EXPLANATION:

please thank to my answer

please MARK me as brainliest

please follow me

Answer:

A moving point X moves in such a way so that it is equidistant (at the same distance) from L1 and L2. What will be its path, or locus? In other words, how will X move given the constraint that it is at the same distance from L1 and L2?

Recall the case of the perpendicular bisector. In that case, we wanted a point to be always equidistant from two fixed points. Here, we want a point to be always equidistant from two fixed lines. There is a difference between the two.

The answer to our current problem is that X must lie on one of the two angle BISECTORS of L1 and L2. It shouldn’t be difficult to understand that there will be two angle bisectors, as the figure below shows:

Now, suppose that X lies on any one of the two angle bisectors. We then need to show that X is equidistant from L1 and L2. In other words, the perpendiculars from X onto L1 and L2 will be of the same length:

We need to show that XY = XZ. Consider the two triangles OXY and OXZ:

1. OX = OX (common side)

2. ∠XOY = ∠XOZ (OX is the angle bisector of ang YOZ)

3. ∠XYO = ∠XZO = 90°

By the ASA criterion, ∆OXY ≡ ∆OXZ, which means that XY = XZ. Thus, X is equidistant from L1 and L2. And this result holds regardless of which of the two angle bisectors X lies on.

To summarize: a point equidistant from two lines must lie on one of the two angle bisectors of the angles formed between the two lines.

Example 1: O is any point inside

Δ

A

B

C

Δ A B C

. Prove that

∠

B

O

C

>

∠

B

A

C

∠ B O C > ∠ B A C

.

Solution: Consider the following figure:

Through A, we have drawn AO, which intersects BC at D. We have also marked various angles in the figure.

Note that

∠

1

∠ 1

is an exterior angle for

Δ

A

O

B

.

Δ A O B .

Clearly,

∠

1

>

∠

3

∠ 1 > ∠ 3

Similarly,

∠

2

>

∠

4

∠ 2 > ∠ 4

Adding these two inequalities,we have:

∠

1

+

∠

2

>

∠

3

+

∠

4

⇒

∠

B

O

C

>

∠

B

A

C

∠1+∠2>∠3+∠4⇒∠BOC>∠BAC

Example 2: The bisector of

∠

A

∠ A

of

Δ

A

B

C

Δ A B C

meets BC at D. If

A

B

>

A

C

,

AB>AC,

prove that

∠

A

D

C

∠ A D C

is acute.

Solution: Consider the following figure, which shows

Δ

A

B

C

Δ A B C

with

A

B

>

A

C

AB>AC

:

Since AB > AC, we have

∠

C

>

∠

B

∠ C > ∠ B

(angle OPPOSITE the larger side will be larger). Now, we will make use of the exterior angle theorem. We note that

∠

A

D

B

=

1

2

∠

A

+

∠

C

∠ADB=12∠A+∠C

∠

A

D

C

=

1

2

∠

A

+

∠

B

∠ADC=12∠A+∠B

Clearly,

∠

A

D

B

>

∠

A

D

C

∠ A D B > ∠ A D C

. But since these two angles form a supplementary pair, this further means that

∠

A

D

B

∠ A D B

must be obtuse and

∠

A

D

C

∠ A D C

must be obtuse. Thus, our proof is complete.

Example 3: This problem consists of four parts, which should be covered in sequence

(a) Let D be any point on side BC of

Δ

A

B

C

Δ A B C

. Prove that at least one of AB or AC must be greater than AD.

(b) Let BC be the greatest side of

Δ

A

B

C

Δ A B C

. Suppose that D and E are points on BC and CA re spectively. Prove that

B

C

≥

D

E

BC≥DE

.

(c) Let BC be the greatest side of

Δ

A

B

C

Δ A B C

, and let E and F be points on AB and AC respectively. Prove that

B

C

≥

E

F

BC≥EF

.

(d) Prove that no straight line segment can be drawn within a triangle which is greater than its greatest side.

Solution:

(a) Consider the following figure:

Let us suppose that AC > AB. This means that

∠

B

>

∠

C

∠ B > ∠ C

. Now, using the exterior angle theorem in

∠

A

B

D

,

∠ A B D ,

we see that

∠

A

D

C

=

∠

B

+

∠

B

A

D

⇒

∠

A

D

C

>

∠

B

∠ADC=∠B+∠BAD⇒∠ADC>∠B

But since

∠

B

>

∠

C

,

∠B>∠C,

we have

∠

A

D

C

>

∠

C

,

∠ADC>∠C,

which means (in

∠

A

D

C

∠ A D C

) that AC > AD.

Similarly, if AB > AC, we can show that AB must be larger than AD. In both cases,at least one of AB or AC must be greater than AD (note that both of them could also possibly be greater than AD).

The proof for the case of AB = AC is left to the reader as an exercise.

(b) The following figure shows a

∠

A

B

C

∠ A B C

with BC as its largest side. We have also marked D and E as arbitrary locations on BC and CA respectively:

We need to prove that

B

C

≥

D

E

,

BC≥DE,

i.e., BC cannot be less than DE. Join A to D:

Consider

Δ

A

D

C

Δ A D C

. Using the result of part-(a), at least one of AD or DC must be greater than DE. Now, we will show that BC is greater than both AD and DC:

BC is obviously larger than DC.

Consider

Δ

A

B

C

Δ A B C

. AD must be smaller than at least one of AB or AC (part-(a)), and both AB and AC are smaller than BC (since BC is largest side). Thus, AD < BC.

Since BC is greater than both AD and DC, and at least one of (AD, DC) is greater than DE, we CONCLUDE that BC > DE.

A special case occurs when D coincides with B and E coincides with C. In this case, BC = DE. Combining the two results, we have:

B

C

≥

D

E

B C ≥ D E

(c) Once again, the following figure shows

Δ

A

B

C

Δ A B C

with BC as the largest side. E and F are points on AB and AC respectively:

We need to show that

B

C

≥

E

F

BC≥EF

. Some of you may say that this visually obvious. EF can never be larger than BC. However, we need to demonstrate this through a logical, step-by-step proof.

Consider

Δ

A

F

B

ΔAFB

. EF must be smaller than at least one of AF or BF. We further note that:

AF < AC < BC, because BC is given to be the largest side.

In

∠

A

B

C

∠ A B C

, BF must be smaller than at least one of (AB, AC).

ANSWER:

SEARCH on GOOGLE PLEASE FOLLOW me

Answer:

10/7

Step-by-step EXPLANATION:

= 20/7+15/7-25/7

= 20+15-25 / 7

= 10/7

Answer:

To convert dag to kg USE the FOLLOWING formula:

1 dag equals 10 kg

So, to convert 416 dag to kg, multiply 10 by 416 i.e.,

416 dag = 10 * 416 kg = 4160 kg

Step-by-step EXPLANATION:

hope it helps u

PLS MARK ME AS BRAINLIST

LET length be x and breadth be y

As said we SHALL SUBTRACT 5 from X and y

X-5 and y-5

Perimeter of rectangle = 2. Length + 2. Breadth

=2(x+5) +2(y+5)

= 2x+10+2y+10

=2(x+y+10)

ANSWER:

ANSWER

Given ∠B=90

∘

,AB=28cm,BC=21cm

Area of shaded region=ar△ABC+area of SEMICIRCLE AC-ar quarter circle BC

=

2

1

×BC×AB+

2

1

×π(

2

AC

)

2

−

4

1

×π(BC)

2

=[

2

1

×21×28]+[

2

1

×

7

22

×

4

1

(

28

2

+21

2

)

2

]−

4

1

×

7

22

×21×21

=294+[

28

11

×1225]−346.5

=294+481.25−346.5=428.75㎠

ANSWER:

4+6+5=15cm is the answer

Answer:

A SECTOR is a region bounded by two RADII of a CIRCLE and the INTERCEPTED arc of the circle.

Step-by-step EXPLANATION:

jzjzjahzhsjsnxbjsksbxjskehzkhdjdhgxuxisjehdiehyfiwsuaasddfgghhfdjdhhshagsgradahsnxbbxasdfgh04akjsnxbxhshhdhdhdhdjdjdhdjdjhdhdjdjdhdhdjdjjdudjdjdjdjdjdjdjdjdjrjrjdjdjdjdjdjdjdududjdjcjxjsoxjjxkdgkdjbxdjjjdjdjdjdjdjdjdjdjdjdjdjjdjdjdjdjdjdjjdjdjdjdjdjdjdjdjdjdjdjjdjdjdjjdjjdjcjjdjcjcjxjcjejxiwjbxhewjnnisldnidjebhbchcjdjdii2nduwn1oeonejfwodkndne8kenebxiwnejdiwkdnieo2nd

cos11° + sin11°/ cos11°-sin11° = TANX

let me TAKE it as ,

COSA +sinA/cosA-sinA = tanx

FOLLOW the ATTACHMENT ...;):)

ANSWER:

Q.1 ANS. 38

Q.2 Ans. RS. 360

Q.3 Ans. Rs. 42

Q.4 Ans. Two more weeks (Not SURE in this QUES)

ANSWER:

5000+6000+9000+7500=27500 is the answer

Answer:

quadrant

Answer. the COORDINATE AXES divide the PLANE into four parts CALLED quadrant

Answer:

OHH U R also in 10th

But SORRY I've came here knlw this ANS only...

ANSWER:

the an_ of this QUE is 12

Step-by-step explanation:

so it's definitely TRUE

Mark me as BRAINLIEST

follow me

ANSWER:

the PROPORTION of 4 and 25 is 10

Answer:

the time and EFFORT to do the same time for me but it seems that they are addressed to get to know that we will have it on your site in this email at the END I was wondering why the time you have the OPPORTUNITY of a COUPLE DAYS and nights and a great night

Step-by-step EXPLANATION:

no TANGENTS INCLUDE at the CONTACT of the CIRCLE in other than 90°

Answer:

PLEASE refer IMAGE for better understanding

Step-by-step explanation:

REF.Image.

As the know that two tangents drawn from an

external point to the circle are EQUAL in length.

so,

AR=AP=2cm

BQ=BP=3cm

RC=QC=4cm

Perimeter =AB+AC+BC

=(AR+RC)+(CQ+BQ)+(AP+BP)

=(2+4)+(4+3)+(2+3)

=6+7+5

=18cm

please mark me as BRAINEST and FOLLOW me

ANSWER:

No

Step-by-step EXPLANATION:

PLEASE MARK me as BRAINLIEST

Step-by-step EXPLANATION:

ueueurusudyeyeyeydteyydydueudigor

ANSWER:

The PRODUCT of (-5)and(+6) is -30

because [-][+] = [-]

Step-by-step explanation:

ANSWER:

54*2+42-3*30-4=15228 is the answer

ANSWER:

E IS THE CORRECT ANSWER JUST CHOOSE E

Step-by-step EXPLANATION:

ANSWER:

3*2=6

and

2*3=6

also

both are same

this is the answer

ANSWER:

in the ratio 2:1

2 is the anticedent...

MARK as brainliest PLS

ANSWER:

Step-by-step EXPLANATION:

hehebeiebeuebdidbdudbdhid

<P>Answer:

Given: Let ABCD is a quadrilateral which circumscribe the circle with CENTER 'O'. The quadrilateral ABCD touches circle at point P , Q , R and S.

To proof: AB+CD = AD+BC

Proof:

AP=AS…(i)

BP=BQ…(ii)

CR=CQ…(III)

DR=DS…(iv)

Adding equation (i),(ii),(iii) and (iv) we get…

AP+BP+CR+DR=AS+BQ+CQ+DS

According to FIGURE [AB=AP+PB, AD=AS+SD, CD=DR+CR, BC=BQ+CQ]

Or, AB+CD=AD+BC (Proved).

In this question a quadrilateral ABCD is drawn to circumscribe a circle.

Hence,

AB+CD=AD+BC

Or, 12+14=AD+15

Or, AD = 11 cm

ANSWER:

THANKS for FREE points

thxxxxxxxxxxxxxxxxxxxxxxx

ANSWER:

WRITE in ENGLISH i am in KERALA

ANSWER:

Question toh DE toh

Step-by-step EXPLANATION:

Question malum HAI kya?

ANSWER:

n3exv3b3bhd3cg4g43b

ANSWER:

YA aapna English mai BOLA HA I kay

Sincere apologies for the inconvenience caused.

2x +3y = 13-----(1)

XY =6-----(2)

8x³ +27y³

= (2x)³ +(3y)³

= (2x+3y)³ - 3*2x*3y(2x+3y) [USING (a+b)³ =

a³+b³+3ab(a+b)]

= (13)³- 18 * 6 *13 [ using (1) and (2)]

= 2197 - 1404

=793

ANSWER:

2*3+3*2+7*30=222 is the answer.

Answer:

10/4

Step-by-step EXPLANATION:

10×9×8×7×6×5×4×3×2×1

=__________________

4×3×2×1

= 10×9×8×7×6×5

Answer:

152....................................

Step-by-step EXPLANATION:

y = 30* VERTICALLY OPPOSITE ANGLE

Answer:

the VOLUME of CUBE increases by 8 times.

Step-by-step EXPLANATION: