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3 years ago

If AB=6x-7 and BC=5x-3 find AC

Mathematics

1 answer:

luda_lava [24]3 years ago

4 0

Answer:

AC= AB+BC

= 6x-7+ 5x-3

= 11x-10

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Solve the following differential equation: y" + y' = 8x^2

PtichkaEL [24]

Answer:

$y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$

Step-by-step explanation:

Let's find a particular solution. We need a function of the form $y= ax^3+bx^2+cx+d$ such that

$y'= 3ax^2+2bx+c$ and

$y''= 6ax+2b$

$y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2$

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a = 8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is $y_p= \frac{8}{3}x^3-8x^2+16x$ .

Now, let's find the solution $y_p$ of the homogeneus equation $y''+y'=0$ with the method of constants coefficients. Let $y=e^{\lambda x}$

$y'=\lambda e^{\lambda x}$

$y''=\lambda^2e^{\lambda x}$

then $\lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0$

$e^{\lambda x}(\lambda +\lambda^2)= 0$

$(\lambda +\lambda^2)= 0$

$\lambda (1+\lambda)= 0$

$\lambda =0$ and $\lambda)= -1$ .

So, $y_h = C_1 + C_2e^{-x}$ and the solution is

$y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$ .

5 0

3 years ago

What is another way to write x2.x4?

puteri [66]

Answer:

If x2.x4 and x2= 2x, x4=4x, while . means multiplication. therefore it can be written as 2x × 4x

3 0

3 years ago

Which expression is equivalent to (x Superscript one-fourth Baseline y Superscript 16 Baseline) Superscript one-half?

Phantasy [73]

The expression that is equal to (x Superscript one-fourth Baseline y Superscript 16 Baseline) is x Superscript one-eighth Baseline y Superscript 8.

<h3>What is an Expression?</h3>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

The expression that is equal to (x Superscript one-fourth Baseline y Superscript 16 Baseline) can be found by simplifying the given algebraic equation,

$(x^{\frac14}y^{16})^{\frac12}\\\\=x^{(\frac14 \times \frac12)} \times y^{(16 \times \frac12)}\\\\= x^{\frac18}y^{8}$

Hence, the expression that is equal to (x Superscript one-fourth Baseline y Superscript 16 Baseline) is x Superscript one-eighth Baseline y Superscript 8.

Learn more about Expression:

brainly.com/question/13947055

#SPJ1

5 0

2 years ago

I need help finding the domain and range, I already know is not a function

FromTheMoon [43]

Answer:

Domain: [-4, 4]

Range: [-3, 5]

Function? No

Step-by-step explanation:

Domain is all x-values that can be inputted in the graph that returns an output.

Range is all y-values that are outputted when <em>x</em> is inputted.

A function has to pass the Vertical Line Test.

6 0

3 years ago

Find the quotient. 64 ÷ 32 =

Viktor [21]

Answer:

64÷32=2.

Step-by-step explanation:

if you add 32+32 it's 64, so that means 32x2=64 and yeahhh

6 0

3 years ago