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2 years ago

3х + 9 = 10x + 9 - 7х Solution? CAN I PLEASEE GET HELPPP ASAPPP!

Mathematics

1 answer:

vampirchik [111]2 years ago

4 0

The answer is equal to each other at my school we call it no solution

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In Science class, Mrs. Olson had 30.6 pounds of metamorphic rock. Mrs. Olson had to divide the rocks into 9 different containers

JulijaS [17]

The answer is 3.4 because 30 can’t be divided equally into 9, reason being that 9 x 30 = 270 which doesn’t make sense, so you have to divide 30 equally into 9 parts then turn it into a ratio.

7 0

3 years ago

Read 2 more answers

Please help me with this

satela [25.4K]

Area of the trapezoid:
$\frac{12.8 + 4.2}{2} (4.9) = 41.65 cm^2$

Area of the rectangle:
4.2 x 3.1 = 13.02 cm^2

Area of each quarter circle:
$\frac{ \pi (3.1)^{2} }{4} = 7.54 cm^2$

We have two quarter circles:
7.54 + 7.54 = 15.08

41.65 + 13.02 + 15.08 = 69.75 cm^2

5 0

2 years ago

black pencils cost 75 naira each and colour pencils cost 105 naira each if 24 mixed pencils cost 2010 naira how many of them wer

KonstantinChe [14]

Answer:

number of black pencils is 17

7 0

2 years ago

9 unit percent change is 23, what is the 19th unit percent cahnge?

vichka [17]

Answer:

42.40705%

Step-by-step explanation:

(1 - (23/100)^(19/9) = 1 + r/100

0.57593 = 1 + r/100

r = 42.40705%

6 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

2 years ago