Question

# Which is greater and

Hint:

### Check whether power of 10 is positive or negative, then move the

decimal places and then compare the values.

## The correct answer is: 0.000006

### Complete step by step solution:

Here can be written as 0.000006

Because of the exponent of 10s term is negative, we must move the decimal 6

places to the left and

can be written as 0.00000002

Because of the exponent of 10s term is negative, we must move the decimal 8

places to the left.

Here it is clear that 0.000006 > 0.00000002

That is,

### Related Questions to study

### Estimate 0.037854921 to the nearest hundredth. Express your answer as a single digit times a power 10 ?

Here the digit at the thousandth place in the given decimal number is 7.

Since 7 > 5, we add 1 to the digit at the hundredth place and remove all the

digits right to it.

So here, we get 0.037854921 rounded to the nearest 100th is 0.04.

This can be expressed as 4 10

^{-2}

### Estimate 0.037854921 to the nearest hundredth. Express your answer as a single digit times a power 10 ?

Here the digit at the thousandth place in the given decimal number is 7.

Since 7 > 5, we add 1 to the digit at the hundredth place and remove all the

digits right to it.

So here, we get 0.037854921 rounded to the nearest 100th is 0.04.

This can be expressed as 4 10

^{-2}

### Sita estimated 30490000000000 as 3 x 10^{8} , What error did she make ?

Here we have 30490000000000 as the number

We have 10 zeros after 3049,

hence we can write 30490000000000 as

### Sita estimated 30490000000000 as 3 x 10^{8} , What error did she make ?

Here we have 30490000000000 as the number

We have 10 zeros after 3049,

hence we can write 30490000000000 as

### Write the standard form for

Here,

Because of the exponent of 10s term is negative, we must move the decimal 8

places to the left.

### Write the standard form for

Here,

Because of the exponent of 10s term is negative, we must move the decimal 8

places to the left.

### Simplify by combining similar terms:

Here we can write, and

This can be written as

So,

Taking common term outside, we have

### Simplify by combining similar terms:

Here we can write, and

This can be written as

So,

Taking common term outside, we have

### Evaluate

Let

∴ (Rationalising the denominator)

Now,

Squaring both sides, we get

### Evaluate

Let

∴ (Rationalising the denominator)

Now,

Squaring both sides, we get

### Simplify 4√12+5√27 - 3√75 +√300

On prime factorization, we can write

and

This can be written as and respectively.

So,

### Simplify 4√12+5√27 - 3√75 +√300

On prime factorization, we can write

and

This can be written as and respectively.

So,

### Simplify √45 - 3√20 +4√5

On prime factorization, we can write

This can be written as and respectively.

So,

### Simplify √45 - 3√20 +4√5

On prime factorization, we can write

This can be written as and respectively.

So,

### Arrange in ascending order of magnitude and

Here we can write,

On taking the LCM of 2,3,6 we have LCM as 6.

So,

Likewise,

Likewise,

Now, we have and

So, here ascending order is 12 < 121 < 125

### Arrange in ascending order of magnitude and

Here we can write,

On taking the LCM of 2,3,6 we have LCM as 6.

So,

Likewise,

Likewise,

Now, we have and

So, here ascending order is 12 < 121 < 125

### The Diagonals of a square ABCD meet at O. Prove that

Aim :- Prove that

Let length of side of square be a

Then ,applying using pythagoras theorem in triangle ADC

We get

As diagonals bisect each other AC = 2OA

where a =AB

We get

Hence proved

### The Diagonals of a square ABCD meet at O. Prove that

Aim :- Prove that

Let length of side of square be a

Then ,applying using pythagoras theorem in triangle ADC

We get

As diagonals bisect each other AC = 2OA

where a =AB

We get

Hence proved

### △ PQR is a right triangle with ∠Q=90^{∘} M is the midpoint of QR. Prove that

As M is mid point of QR then, QM = ½ RQ

We get —Eq1

Applying Pythagoras theorem For triangle PQM

—Eq2

Substitute Eq2 in Eq1

We get

Hence proved

### △ PQR is a right triangle with ∠Q=90^{∘} M is the midpoint of QR. Prove that

As M is mid point of QR then, QM = ½ RQ

We get —Eq1

Applying Pythagoras theorem For triangle PQM

—Eq2

Substitute Eq2 in Eq1

We get

Hence proved

### In Triangle . Prove that

Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get

— Eq1

Applying pythagoras theorem in Δ ABD ,We get

Substitute Eq1 in Eq2 ,

Hence proved

### In Triangle . Prove that

Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get

— Eq1

Applying pythagoras theorem in Δ ABD ,We get

Substitute Eq1 in Eq2 ,

Hence proved

### In and prove that

Aim :- Prove that

Hint :- using inverse of pythagoras theorem, we get right angle at Q,

Now using angles prove the similarity of QMR and PMQ triangles. And find the

ratios to get the desired property

Explanation(proof ) :-

Given ,

Here ,PR is hypotenuse and angle Q is right angle.

Then

We know

By AA similarity

By similarity we get

Hence proved.

### In and prove that

Aim :- Prove that

Hint :- using inverse of pythagoras theorem, we get right angle at Q,

Now using angles prove the similarity of QMR and PMQ triangles. And find the

ratios to get the desired property

Explanation(proof ) :-

Given ,

Here ,PR is hypotenuse and angle Q is right angle.

Then

We know

By AA similarity

By similarity we get

Hence proved.

### In the figure,then find the value of x and y.

Step 1:-Find ∠QSR

We know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.

∠QSR +∠SQR = ∠QRT

∠QSR +28° = 65°

∠QSR = 65° -28° = 37°

Step 2:- Find value of x

PQ || SR and QS intersects both the lines

We get angle x = ∠QSR ( alternate interior angles)

x =37°

Step 3:- Find value of y

In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)

90°+37° +y =180 °

y = 180-127

y = 53 °

### In the figure,then find the value of x and y.

Step 1:-Find ∠QSR

We know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.

∠QSR +∠SQR = ∠QRT

∠QSR +28° = 65°

∠QSR = 65° -28° = 37°

Step 2:- Find value of x

PQ || SR and QS intersects both the lines

We get angle x = ∠QSR ( alternate interior angles)

x =37°

Step 3:- Find value of y

In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)

90°+37° +y =180 °

y = 180-127

y = 53 °

### Arrange in descending order of magnitude and

Here we can write, and

On taking the LCM of 3,6,9 we have LCM as 18.

So,

Likewise,

Likewise,

Now, we have and

So, here descending order is 64 > 27 > 16

### Arrange in descending order of magnitude and

Here we can write, and

On taking the LCM of 3,6,9 we have LCM as 18.

So,

Likewise,

Likewise,

Now, we have and

So, here descending order is 64 > 27 > 16

### In △ABC, AD⊥BC Also, Prove that

As BD = 3 CD ;We get BC = BD +CD = 4 CD

Applying pythagoras theorem in ΔACD ,We get

Applying pythagoras theorem in ΔABD ,We get

—Eq2

Substitute Eq2 in Eq1 ,

As BD = 3 CD ;We get

Multiplying with 2 on both sides

As BC = 4 CD

∴

Hence proved

### In △ABC, AD⊥BC Also, Prove that

As BD = 3 CD ;We get BC = BD +CD = 4 CD

Applying pythagoras theorem in ΔACD ,We get

Applying pythagoras theorem in ΔABD ,We get

—Eq2

Substitute Eq2 in Eq1 ,

As BD = 3 CD ;We get

Multiplying with 2 on both sides

As BC = 4 CD

∴

Hence proved