CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

Brain tumours: Abnormal growth of tissue within the brain is termed as the brain tumour.
Neurodegenerative Diseases: A neurodegenerative disease is also called as the Degenerative Nerve Diseases.
Meningitis, Encephalitis, Seizures, Trauma, Hydrocephalus, Brain abscess, Tumors are the list of most common brain diseases.

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find the area of the quadrilateral abcd in which ab =50m bc=60m cd=30m da=90m and bd =70m

Posted by Cbz Hero 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Cbz Hero 3 years, 10 months ago

Thanks

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Gaurav Seth 3 years, 10 months ago

Step-by-step explanation:

in triangle BCD,

semiperimeter (s) = BC + CD + BD/2

s = 60 + 30 + 70 /2

s = 160/2

s = 80

a = root[80 x (80-60) (80-30) (80 - 70)]

a = root[80 x 20 x 50 x 10]

a = root[ 10 x 4 x 2 x 2 x 10 x 5 x 10 x 10]

a = 10 x 4 x 10 root(5)

a = 400xroot(5)

a = 894.4 (approx)

In triangle ABD,

semiperimeter (s) = AB + BD + DA/2

s = 50 + 70 + 90/2

s = 210/2

s = 105

a = root[105 x (105- 50) (105-90) (105 - 70)]

a = root[105 x 55 x 15 x 35]

a = root[ 5 x 7 x 3 x 5 x 11 x 5 x 3 x 7 x 5]

a = 25 x 7 x 3 root(11)

a = 525 * root(11)

a = 1741.2 (approx)

Total area = 894.4 + 1741.2

= 2635.6 cm2(approx

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Is square root 5 is a rational number

Posted by Dibya Ranjan Nayak 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Sanjay Kumar 3 years, 10 months ago

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Gaurav Seth 3 years, 10 months ago

Let us assume that √5 is a rational number.

So it t can be expressed in the form p/q where p,q are co-prime integers and q≠0
⇒√5=p/q

On squaring both the sides we get,
⇒5=p²/q²
⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p
p is a multiple of 5
⇒p=5m
⇒p²=25m² ————-(ii)
From equations (i) and (ii), we get,
5q²=25m²
⇒q²=5m²
⇒q² is a multiple of 5
⇒q is a multiple of 5
Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number

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In the adjoining figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. Chapter - 13 (13.2) question 10

Posted by Priya Rana 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 3 years, 10 months ago

In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

<hr />

2r = 20 cm ⇒ r = 10 cm h = 30 cm
Cloth required = 2πr(h + 2.5 + 2.5)

= 2200 cm².

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Find the angle whose complement is equal to the angle itself.

Posted by Vèñkãťêšh ? 3 years, 10 months ago

CBSE > Class 09 > Mathematics

3 answers

Мυѕкαи ? 3 years, 10 months ago

45° dear friend

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Yogita Ingle 3 years, 10 months ago

Let the measure of an angle be x, then the, measure of its complement is also x.

we know that the sum of the measures of complementary angles is 90˚.

Therefore, x + x = 90˚

2x = 90˚

x = 90/2 = 45˚

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Sigatapu Durga 3 years, 10 months ago

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Factorise : p -81p

Posted by Riya Gujjar 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Charu Nethara 3 years, 10 months ago

-80p

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Under root 10×under root 15

Posted by Pawan Kumar 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Anushka Sharma 3 years, 10 months ago

√10 ×√15= √5×2×√5×3 = √5×5×2×3 =5√6=

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Charu Nethara 3 years, 10 months ago

The answer is under root 150

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Formula of probability count head

Posted by Amruta Shinde 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Sanjay Kumar 3 years, 10 months ago

Probability P(E)= 1/2

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Vandan Dev Barnwal 3 years, 10 months ago

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Long division method 5examples

Posted by Ayush Kumar 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 3 years, 10 months ago

STEP 1: Focus on the leftmost terms of both the dividend and divisor.

STEP 2: Divide the leftmost term of the dividend by the leftmost term of the divisor.

STEP 3: Place the partial answer on top.

</figure>

STEP 4: Use that partial answer, x², to multiply into the divisor (3x−2).

STEP 5: Place their product under the dividend. Make sure to align them by similar terms.

</figure>

STEP 6: Perform subtraction by alternating the signs of the bottom polynomial.

</figure>

STEP 7: Proceed with regular addition vertically. Again the first column cancels each other out. Looks like a pattern to me!

STEP 8: Carry down the next adjacent “unused” term of the dividend

</figure>

STEP 9: Take the leftmost term of the bottom polynomial, and divide by the leftmost term of the divisor.

STEP 10: Place the answer on top, as usual.

</figure>

STEP 11: Okay, perform another multiplication by the partial answer 2x and divisor (3x−2). Bring the product below.

</figure>

STEP 12: Perform subtraction by switching signs and proceed with normal addition.

STEP 13: Carry down the last unused term of the dividend. We’re almost there!

</figure>

STEP 14: We are going up one more time. Divide the leading term of the bottom polynomial by the leading term of the divisor. Place the answer up there!

</figure>

STEP 15: This is our “last trip” going down so we distribute the partial answer −1 by the divisor (3x−2), and placing the product “downstairs”.

</figure>

STEP 16: Finish this off by subtraction leaving as with a remainder of −7.

</figure>

STEP 17: Write the final answer in the following form.

</figure> <figure>

</figure> <hr />

Example 3: Divide using the long division method

</figure>

Solution: If you observe the dividend, it is missing some powers of variable x which are x³ and x². I need to insert zero coefficients as placeholders for missing powers of the variable. This is a critical part to correctly apply the procedures in long division.

So I rewrite the original problem as . Now all x‘s are accounted for!

</figure>

STEP 1: Focus on the leading terms inside and outside the division symbol.

STEP 2: Divide the first term of the dividend by the first term of the divisor.

STEP 3: Position the partial answer on top.

</figure>

STEP 4: Use that partial answer placed on top, 3x² to distribute into the divisor (x + 1).

STEP 5: Put the result under the dividend. Make sure to align them by similar terms.

</figure>

STEP 6: Subtract them together by making sure to switch the signs of the bottom terms before adding.

</figure>

STEP 7: Carry down the next unused term of the dividend.

</figure>

STEP 8: Looking at the bottom polynomial, −3x³+ 0x², use the leading term −3x³ and divide it by the leading term of the divisor, x. Put the answer above the division symbol.

</figure>

STEP 9: Multiply the answer you got previously, −3x³, and distribute into the divisor (x + 1).

STEP 10: Place the answer below then perform subtraction.

</figure>

STEP 11: Bring down the next adjacent term of the dividend

</figure>

STEP 12: Go up again by dividing the leading term below by the leading term of the divisor.

</figure>

STEP 13: Go down by distributing the answer in partial quotient into the divisor, followed by subtraction.

I believe the pattern makes sense now. Yes?

</figure>

STEP 14: Carry down the last term of the dividend.

</figure>

STEP 15: Go up again while performing division.

</figure>

STEP 16: Go down again while performing multiplication.

</figure>

STEP 17: Do the final subtraction, and we are done! The remainder is equal to 20.

</figure>

STEP 18: The final answer is

</figure> <figure>

</figure> <hr />

Example 4: Divide the given polynomial using long division method

</figure>

Solution: The dividend is obviously missing a lot of variable x. That means I need to insert zero coefficients in every missing power of the variable.

I need to rewrite the problem this way to include all exponents of x in descending order:

</figure>

Remember the Main Steps in Long Division:

When going up, we divide
When going down, we distribute
Subtract
Carry down
Repeat the process until done

Verify if the steps are being applied correctly in the example below.

</figure>

So the final answer is

</figure>

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What is difference between polynomial and algebra

Posted by Uttam Gautam 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Shubham Naik 3 years, 10 months ago

Nothing????

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If the non parallel sides of a trapezium are equal prove it is acyclix

Posted by Mohammed Basil 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Shubham Naik 3 years, 10 months ago

Shut up man ???

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If diagonal of a cyclic qu

Posted by Mohammed Basil 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 3 years, 10 months ago

If the diagonals of a cyclic quadrilateral are diameters of the circle through the opposite vertices of the quadrilateral. Prove that the quadrilateral is a rectangle.

Answer

Here, ABCD is a cyclic quadrilateral in which AC and BD are diameters .

Since AC is a diameter.

∴ ∠ABC = ∠ADC = 90°

[∵ angle of a semicircle = 90°]

Also, BD is a diameter

∴∠BAD = ∠BCD = 90°

[∵ angle of a semicircle = 90°]

Now, all the angles of a cyclic quadrilateral ABCD are 90° each.

Hence, ABCD is a rectangle.

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I want to ask 13.1 question 3.. please explain

Posted by Saloni Gupta 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Chaitanya Goravar 3 years, 10 months ago

Given: perimeter of rectangular wall =2(l+b)=250m (1) Now area of the four walls of the room =total cost to paint walls of the room ________________________________ Cost to paint 1m² of the wall =15000 = 1500 m² (2) ________ 10 Therefore, area of the four walls = lateral surface area 2(bh+hl) = 2h(b+l) =1500 250×h = 1500 (from equation 1 & 2) h = 1500 ________ =6m 250 Hence, required height of the hall is 6m.

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Kirti and sarita two students of class x of a school , together contributed Rs100 towards prime minister's relief fund to help earthquake victims. Write a linear equation which satisfies this data. U may take their contribution as Rs x and Rs y.

Posted by Parul Saini 3 years, 10 months ago

CBSE > Class 09 > Mathematics

4 answers

Sanjay Kumar 3 years, 10 months ago

x+y=100

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Sanskrati Kesharwani 3 years, 10 months ago

Total money = ₹100 Contribution of Kirti = ₹ x Contribution of Sarita = ₹ y According to the question- ₹ x + ₹ y = ₹ 100

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Abid Farhan {{Devil}} 3 years, 10 months ago

X + Y =100

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Bandana Kundu 3 years, 10 months ago

X + Y = 100

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The angles of quadrilateral are in the ratio 3:5:9:13. find all the angles of the quadrilateral.

Posted by Prachi - 3 years, 10 months ago

CBSE > Class 09 > Mathematics

5 answers

Prachi - 3 years, 10 months ago

which subham...

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Priya Yadav 3 years, 10 months ago

Hello

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Cbz Hero 3 years, 10 months ago

It's me shubham

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Yukti Verma 3 years, 10 months ago

Let them x. Then all angle will be 3x,5x,9x and 13x So, 3x +5x +9x+ 13x = 360 30x = 360 X = 12 degree So, required angles are 36,60,108 and 156 degree Thank you

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Prachi - 3 years, 10 months ago

pls can anyone reply

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Define Modal verbs and Auxiliary verbs ?

Posted by Neeraja D 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Yogita Ingle 3 years, 10 months ago

Modals (also called modal verbs, modal auxiliary verbs, modal auxiliaries) are special verbs which behave irregularly in English. They are different from normal verbs like “work, play, visit…” They give additional information about the function of the main verb that follows it. They have a great variety of communicative functions.

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Priya Dagar 3 years, 10 months ago

Modal verbs are auxiliary verbs (also called helping verbs) like can, will, could, shall, must, would, might, and should. ... Modal verbs add meaning to the main verb in a sentence by expressing possibility, ability, permission, or obligation. You must turn in your assignment on time. He might be the love of my life.

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What is square of 65

Posted by Karan K U M A R 3 years, 10 months ago

CBSE > Class 09 > Mathematics

4 answers

Shailee Mathpal 3 years, 10 months ago

4225

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Govind Sharma 3 years, 10 months ago

The square of 65 is 4225

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Priya Dagar 3 years, 10 months ago

The square of 65 is 4225

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Surjit Kaur 3 years, 10 months ago

The square of 65 is 4225

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Draw a line pq=7.5 and divide it in 4 equal parts

Posted by Siddhant Agarwal 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Priya Dagar 3 years, 10 months ago

ANSWER In the figure, Step-1 Draw a line segment AB=7.5 cm According to the question- 2x+3x=7.5 cm x=1.5 Step-2 Mark 5 arcs of equal length of 1.5 cm Hence, AC=2×1.5⇒3.0 cm BC=3×1.5⇒4.5 cm. Hence, this is the answer. solution

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Definition of polynomial

Posted by Aarti Raghuwanshi 3 years, 10 months ago

CBSE > Class 09 > Mathematics

3 answers

Priya Dagar 3 years, 10 months ago

Polynomials are expressions with one or more terms with a non-zero coefficient. A polynomial can have more than one term. In the polynomial, each expression in it is called a term.Polynomials are expressions with one or more terms with a non-zero coefficient.

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Queen 14 3 years, 10 months ago

Polynomials are expressions with one or more terms with a non-zero coefficient.

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Yogita Ingle 3 years, 10 months ago

Polynomials are expressions with one or more terms with a non-zero coefficient. A polynomial can have more than one term. In the polynomial, each expression in it is called a term.

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Actually vo mujhe 7.1 ka Question 8 clear nhi ho rha hai

Posted by Misba Aarfi 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Priya Dagar 3 years, 10 months ago

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. (See figure) Show that: (i) AMC BMD (ii) DBC is a right angle. (iii) DBC ACB (iv) CM = AB Ans. (i) In AMC and BMD, AM = BM [AB is the mid-point of AB] AMC = BMD [Vertically opposite angles] CM = DM [Given] AMC BMD [By SAS congruency] ACM = BDM ……….(i) CAM = DBM and AC = BD [By C.P.C.T.] (ii) For two lines AC and DB and transversal DC, we have, ACD = BDC [Alternate angles] AC DB Now for parallel lines AC and DB and for transversal BC. DBC = ACB [Alternate angles] ……….(ii) But ABC is a right angled triangle, right angled at C. ACB = ……….(iii) Therefore DBC = [Using eq. (ii) and (iii)] DBC is a right angle. (iii) Now in DBC and ABC, DB = AC [Proved in part (i)] DBC = ACB = [Proved in part (ii)] BC = BC [Common] DBC ACB [By SAS congruency] (iv) Since DBC ACB [Proved above] DC = AB AM + CM = AB CM + CM = AB [ DM = CM] 2CM = AB CM = AB

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If P (a) =0 then a is called...... Of polynomial

Posted by Aarti Raghuwanshi 3 years, 10 months ago

CBSE > Class 09 > Mathematics