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3 years ago

(-4) (?) = -12 What is the value of the (?)

Mathematics

2 answers:

ehidna [41]3 years ago

5 0

Answer:

(-4) (3)= -12

Step-by-step explanation:

because negative times a positive =a negative

Crazy boy [7]3 years ago

4 0

? = 3

-12/-4 = 3
you can check your answer by plugging it into the variable

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6+n/60=15/90 How do I dolce this

Lilit [14]

$6+\frac{n}{60}=\frac{15}{90}\\\\6+\frac{n}{60}=\frac{1}{6}\ \ \ \ \ |multiply\ both\ sides\ by\ 60\\\\360+n=10\ \ \ \ \ \ \ |subtract\ 360\ from\ both\ sides\\\\\boxed{n=-350}$

8 0

3 years ago

Pls help me, its not hard

natulia [17]

Answer:

1. y = -x + 3

2. y = x - 3

3. y = 3x + 1

4. y = -3x - 3

Step-by-step explanation:

Hello! sorry, I just saw your message on the question I answered before asking for help so here you go, these are the answers, hope this helps.

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3 0

3 years ago

Geometry homework!!!HELP ME PLEASE!!

Lady_Fox [76]

f= (45*23)/35

k=(35*14)/23

5 0

3 years ago

N/5 + -6 = 9 What is n ?

Tatiana [17]

Answer: 75

work: n/5 + -6 = 9
n/5 = 9+ 6
n/5 = 15
n = 15 x 5
n= 75

4 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago