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3 years ago

*I thought of a number, added 18, then multiplied the result by 8. I got 16. What was my number?

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

4 0

Answer:

-16

Step-by-step explanation:

Represent the number by n.

Then:

8(n + 18) = 16

Performing the indicated multiplication, we get:

8n + 144 = 16

Subtracting 144 from both sides results in 8n = -128.

Dividing both sides by 8: n = -128/8 = -16

The number was -16.

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krek1111 [17]

If BC and XY are similar and BC=24 with a ratio of 3:2, then divide 24 by 3 so,

24/3 = 8

8 is the stem unit meaning

3•8= 24

then 2•8= 16

The answer is 16.

3 0

3 years ago

The answer to all of these problems

-BARSIC- [3]

1) area of a cicular pool
area of a circle=πr^2 d=2.5 then r=2.5/2
A=3.14 *(2.5/2)^2
A=4.906m^2 A=4.9m^2

2) arae of a coin
<span>(if i read it right d=15mm?)
</span>A=πr^2 d=15 then r=15/2
A=3.14*(15/2)^2 =176.6mm^2

Rest all are same problems u try to solve it

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3 years ago

What is the slope of the equation -4x-7y=-14

castortr0y [4]

The answer to your question is precisely 99

3 0

3 years ago

F(x) = 3x^4 is even or odd?

lara31 [8.8K]

Answer:

Step-by-step explanation:

it's even because f(-x)=f(x)

3 0

3 years ago

Read 2 more answers

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