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3 years ago

8

A statement that is true both forwards and backwards is called a?

2 answers:

wel3 years ago

8 0

A statement that is true both forwards and backwards. That is the original statement<span> and its converse are </span>both true<span>.</span>

spin [16.1K]3 years ago

6 0

A Bi-Conditional Statement - <span>A statement that is true both forwards and backwards. That is the original statement and its converse are both true</span>-

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Could someone solve this for me.

Studentka2010 [4]

The answer is 7.6 units

5 0

2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

Callie has a new kitten. They can weigh 3 pounds less than half the weight of Callie‘s cat. Together the cat and kitten weigh 18

gavmur [86]

Answer:

The weight of cat is <u>14 pounds</u> and the weight of kitten is <u>4 pounds</u>.

Step-by-step explanation:

Given:

Callie has a new kitten. It can weigh 3 pounds less than half the weight of Callie‘s cat. Together the cat and kitten weigh 18 pounds.

Now, to find the weight of each animal:

Let the cat's weight be $x.$

And the kitten weight = $\frac{x}{2} -3$

Total weight of cat and kitten = 18 pounds.

Now, to set an equation to get the weight of each animal:

$(x)+(\frac{x}{2}-3)=18$

$x+\frac{x}{2} -3=18$

$\frac{2x+x-6}{2} =18$

$\frac{3x-6}{2} =18$

<em>Multiplying both sides by 2 we get:</em>

<em /> $3x-6=36$ <em />

<em>Adding both sides by 6 we get:</em>

<em /> $3x=42$ <em />

<em>Dividing both sides by 3 we get:</em>

<em /> $x=14\ pounds.$ <em />

<em>The weight of cat = 14 pounds.</em>

Substituting the value of $x$ to get the kitten's weight:

$\frac{x}{2} -3\\\\=\frac{14}{2}-3\\\\=7-3\\\\=4.$

<em>The kitten's weight = 4 pounds.</em>

Therefore, the weight of cat is 14 pounds and the weight of kitten is 4 pounds.

7 0

3 years ago

PLEASE HELP ME I BEG

Kay [80]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

• 1 + cot²x = csc²x

Consider the left side

$csc^{4}$ x - $cot^{4}$ x

Factorise as a difference of squares

= (csc²x - cot²x)(csc²x + cot²x)

= (1 + cot²x - cot²x)(csc²x + cot²x)

= csc²x + cot²x

= csc²x + csc²x - 1

= 2csc²x - 1 = right side ⇒ verified

8 0

3 years ago

A different children's party cost £225.50 How many children were at the party?

Alecsey [184]

Hi, I think you forgot to write a part of the problem, could you write the rest in the comments please. I would be happy to help you

8 0

3 years ago