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3 years ago

What are the excluded values?

Mathematics

1 answer:

OverLord2011 [107]3 years ago

7 0

Answer:

c. x = -9.

Step-by-step explanation:

The excluded value is the value of x which gives a value of 0 to the denominator, so that value would be x = -9.

Note: (x - 7) / 0 is undefined.

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Please help with this math question

Ymorist [56]

Area of the rectangle minus area of 2 circles

Area of rectangle is 8x16
Area of circle is pi•r^2 = 3.14 x 4^2= 3.14•16

Answer is:
128 - 2(3.14•16) square inches

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3 years ago

Math Tutor Pay: $11.44 per hour

slavikrds [6]

Answer:

<h2>Per week: $1041.04</h2><h2>Per month: $4610.32</h2><h2>Per year: 55323.84</h2>

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3 years ago

Read 2 more answers

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

frozen [14]

Answer:

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner $I_{corner}$ is:

$I_{corner} = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_{corner} = \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx$

$I_{corner} = \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx$

$I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}$

$I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]$

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is $I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

7 0

3 years ago

Please help I’m so close to passing math

kirill115 [55]

Answer:

$0.14 per ounce of salsa

Step-by-step explanation:

2.80/20

Move decimal to make the number a whole number

(Have to move the decimal the same amount of spaces)

28/200

0.14

Hope this helps. Have a nice day

8 0

2 years ago

Read 2 more answers

Verify a(b-c)=ab-ac for a=1.6;b=1/-2;& c=-5/-7

harina [27]

Given:

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

To verify:

$a(b-c)=ab-ac$ for the given values.

Solution:

We have,

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

We need to verify $a(b-c)=ab-ac$ .

Taking left hand side, we get

$a(b-c)=1.6\left(\dfrac{1}{-2}-\dfrac{-5}{-7}\right)$

$a(b-c)=1.6\left(-\dfrac{1}{2}-\dfrac{5}{7}\right)$

Taking LCM, we get

$a(b-c)=1.6\left(\dfrac{-7-10}{14}\right)$

$a(b-c)=\dfrac{16}{10}\left(\dfrac{-17}{14}\right)$

$a(b-c)=\dfrac{8}{5}\left(\dfrac{-17}{14}\right)$

$a(b-c)=-\dfrac{68}{35}\right)$

Taking right hand side, we get

$ab-ac=1.6\times \dfrac{1}{-2}-1.6\times \dfrac{-5}{-7}$

$ab-ac=-\dfrac{16}{20}-\dfrac{8}{7}$

$ab-ac=-\dfrac{4}{5}-\dfrac{8}{7}$

Taking LCM, we get

$ab-ac=\dfrac{-28-40}{35}$

$ab-ac=\dfrac{-68}{35}$

Now,

$LHS=RHS$

Hence proved.

7 0

2 years ago