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3 years ago

Please please please help meeeee Factor. 9x2−49y2

Mathematics

2 answers:

denis-greek [22]3 years ago

7 0

9x^2 - 49y^2

(3x - 7y)(3x + 7y)

Galina-37 [17]3 years ago

5 0

This Is how you factor 9x2-49y2 ................. (3x+7y)(3x-7y)

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In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random a

Masteriza [31]

The solution to the answer is as follows:

0votesanswered Aug 8, 2015 by Shashi Kumar Evangelist (3,082 points)total coins 30
1 rupee coins 20
50 paise coins 10

1 rupee coin probability 20C11
11 coin are picked 30C11
ans should be = 20C11/30C11 = 2/3
I hope my answer has come to your help. God bless and have a nice day ahead!

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4 years ago

Given a mean of 52.3 and a standard deviation of 1.8, what is the z-score of the value 51.8 rounded to the nearest tenth?

postnew [5]

The z-score of a certain data point in the given statistical data is calculated through the equation,
z-score = (x - m) / sd
where x is the data point, m is the mean and sd is the standard deviation.
Substituting the known values,
z-score = (51.8 - 52.3) / 1.8
z-score = -0.277
Rounding the answer to the nearest tenth will give an answer of -0.3.

8 0

3 years ago

What is the volume of corn held in this cone-shaped grain silo? Use 3.14 for π and round to the nearest cubic foot. Enter your a

babymother [125]

Answer:

V = 301 ft^3

Step-by-step explanation:

The volume of a cone is given by

V =1/3 pi r^2 h

We know pi = 3.14 r =6 and

We need to determine h

We can find h from the Pythagorean theorem where 6 is a leg and 10 is the hypotenuse

h^2 +6^2 = 10^2

h^2 +36 = 100

h^2 = 64

Taking the square root of each side

h = 8

V =1/3 pi r^2 h

V = 1/3 (3.14) (6)^2 * 8

V = 301.44 ft^3

Rounding to the nearest cubic foot

V = 301 ft^3

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4 years ago

Read 2 more answers

The product of a number and 3 is 87 what is the number

IrinaK [193]

The awnser would be 29

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3 years ago

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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$