All categories

3 years ago

The sum of four times a number and negative three is twelve. What is the number?

Mathematics

2 answers:

madam [21]3 years ago

8 0

Answer:

answer is 3 and 3/4

neonofarm [45]3 years ago

3 0

4x-3=12
add 3
4x=15
divide 4
x=15/4
x=12/4+3/4
x=3+3/4
x=3 and 3/4

answer is 3 and 3/4

You might be interested in

What number should be placed in the box to help complete the division calculation. Pls help

ArbitrLikvidat [17]

Answer:

I believe it would be 4

3 0

3 years ago

Plz help this is due today AND NO LINKS OR GROSS PICTURES OR I REPORT

Zina [86]

Answer:

I think its A

Step-by-step explanation:

3 0

3 years ago

Is 42(x + 14y) equivalent to 42x - 56y

g100num [7]

Answer:

No its not

Step-by-step explanation:

When 42(x + 14y) is simplified:

42(x + 14y)

Expand and remove the brackets. Remember brackets mean to multiply.

42*x

42x

42*14y

588y

42x+588y

Therefore 42x-56 is NOT EQUIVALENT to 42(x + 14y) but is equivalent to 42+588y

3 0

4 years ago

Could you help me find X,

lisov135 [29]

when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one. Check the picture below.

$\cfrac{4}{x}~~ = ~~\cfrac{x}{9}\implies 36=x^2\implies \sqrt{36}=x\implies 6=x$

3 0

2 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

4 years ago

Read 2 more answers