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3 years ago

Please help me

Mathematics

2 answers:

Lunna [17]3 years ago

5 0

Answer:

1) y=3.5+(5•x)

2) 103.5

3) No, because 5•20 is 100 and 5•10 is 50, BUT with both you still have to add 3.5 no matter what because of the $3.50 to pick up the rider

Charra [1.4K]3 years ago

3 0

Answer:

im gonna go with c

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Ilia_Sergeevich [38]

Your answer is
${x}^{2} \div {2} {z}^{9}$
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Ms hills travels 2400 miles at a rate of 400 miles per day to visit her grandchild . Write and solve an equation to find the num

Naya [18.7K]

It Would Take her 6 days

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3 years ago

What is the r-value of the following data to three decimal places

Liula [17]

Answer:

option B

$-0.953$

Step-by-step explanation:

we know that

The<u> correlation coefficient</u> r measures the direction and strength of a linear relationship.The value of r is always between $+1$ and $-1$

Using a Excel tool (Correl function)

see the attached table

The value of r is $-0.953$

7 0

3 years ago

Read 2 more answers

Halla la suma de 3, ocho veces un número, y tres veces el mismo número

prohojiy [21]

Answer:

huh?

Step-by-step explanation:

6 0

3 years ago

Consider the differential equation x2y′′ − 9xy′ + 24y = 0; x4, x6, (0, [infinity]). Verify that the given functions form a funda

pantera1 [17]

Answer:

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

Step-by-step explanation:

Given equation is

$x^2y'' - 9xy+24y=0$

This Euler Cauchy type differential equation.

So, we can let

$y=x^m$

Differentiate with respect to x

$y'= mx^{m-1}$

Again differentiate with respect to x

$y''= m(m-1)x^{m-2}$

Putting the value of y, y' and y'' in the differential equation

$x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0$

$\Rightarrow m(m-1)x^m-9mx^m+24x^m=0$

$\Rightarrow m^2-m-9m+24=0$

⇒m²-10m +24=0

⇒m²-6m -4m+24=0

⇒m(m-6)-4(m-6)=0

⇒(m-6)(m-4)=0

⇒m = 6,4

Therefore the auxiliary equation has two distinct and unequal root.

The general solution of this equation is

$y_1(x)=x^4$

and

$y_2(x)=x^6$

First we compute the Wronskian

$W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|$

$= \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|$

=x⁴×6x⁵- x⁶×4x³

=6x⁹-4x⁹

=2x⁹

≠0

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

5 0

3 years ago