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2 years ago

0.006772 to 1 significant number

Mathematics

2 answers:

Pani-rosa [81]2 years ago

6 0

Answer:

0.007

Step-by-step explanation:

Rounding off 0.006772 to 1 significant figures:

=> 0.007

There is only 1 significant figure in this , since the zeroes on the left are not counted as significant figures.

sweet-ann [11.9K]2 years ago

3 0

Answer:

0.006772

If the last dropping digit is less than 5 then it will be ignored

0.00677

if the last dropping digit is greater than 5 than the the last retained digit increases by 1

0.0068

if the last dropping digit is greater than 5 than the the last retained digit increases by 1

0.007

if the last dropping digit is greater than 5 than the the last retained digit increases by 1

0.01

if the last digit is less than 5 so it will be ignored

0.0 is significant figure because zero to the right of decimal point are significant

Step-by-step explanation:

i hope this will help you :)

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A sum amount <br>to rupees 34476 in two and a half years at 4% CI. the sum is

Kruka [31]

Given:

Amount = Rs. 34476

Rate of compound interest = 4%

Time = $2\dfrac{1}{2}=2.5$ years

To find:

The principal value.

Solution:

Formula for amount is

$A=P\left(1+\dfrac{r}{100}\right)^t$

Where, P is principal value, r is rate of interest and t is time in years.

Putting the given values, we get

$34476=P\left(1+\dfrac{4}{100}\right)^{2.5}$

$34476=P\left(1+0.04\right)^{2.5}$

$34476=P\left(1.04\right)^{2.5}$

$\dfrac{34476}{\left(1.04\right)^{2.5}}=P$

Now,

$P=31256.0090$

$P\approx 31256$

Therefore, the value of sum or principal value is Rs.31256.

6 0

2 years ago

A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32

zavuch27 [327]

$Hypotenuse =39cm$ of right angle triangle! .

<u>Step-by-step explanation:</u>

Here we have , A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the length of the other leg is 25 centimeters. The student incorrectly says that the length of the hypotenuse is 7.5 centimeters. We need to find:

We have , One length as 32 cm and other as 25 cm , as

$Perpendicular = 32cm\\Base=25cm$

By Pythagoras Theorem ,

⇒ $Hypotenuse^2 = Perpendicular^2+Base^2$

⇒ $Hypotenuse^2 =32^2+22^2$

⇒ $\sqrt{Hypotenuse^2} =\sqrt{1024+484}$

⇒ $Hypotenuse =39cm$

So , $Hypotenuse =39cm$ of right angle triangle! .

4 0

3 years ago

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Simplify the expression. 34 • 3

bekas [8.4K]

Answer:

102

Step-by-step explanation:

34*3=102

the dot in the middle is just another way to say that it is multiplying the equation so it just mean that your are multiplying 34 by 3 which is 102

3 0

3 years ago

PLEASE HURRY

s2008m [1.1K]

Answer:

1.25p

Step-by-step explanation:

p = amount raised last year

amount raised this year = p +(25% of p)

=1 p+0.25p

=p(1+0.25)

=1.25p

8 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

2 years ago

Read 2 more answers