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3 years ago

19 + x/2 = 4 How do I do this

Mathematics

1 answer:

GuDViN [60]3 years ago

6 0

Answer: x = -30

Steps: 19 + x/2 = 4

Subtract 19 from both sides: 19 + x/2 - 19 = 4 - 19

Simplify: x/2 = 15

Multiply both sides by two: x/2 (2) = -15 (2)

Simplify: x = -30

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Find the length of AB

ryzh [129]

Answer:

use the formula given in the pic and you will get the answer

3 0

3 years ago

It takes Bruce 12 h to complete his training. It takes Oliver approximately 2 days, 2 h, and 15 min to complete his training. Ho

Bond [772]

Answer:

Oliver completed his training work in 50.25 hours

Step-by-step explanation:

we know that

$1\ day= 24\ hours$

$1\ hour=60\ minutes$

we have

$2\ days, 2\ h,\ and\ 15\ min$

Convert days to hours

$2\ days=2(24)=48\ hours$

Convert minutes to hour

$15\ min=15(1/60)=1/4=0.25\ h$

$2\ days, 2\ h,\ and\ 15\ min=(48)+(2)+(0.25)=50.25\ h$

therefore

Oliver completed his training work in 50.25 hours

4 0

3 years ago

Solve equation for the specified variable dx +t=10 , solve for x

slamgirl [31]

Dx+t=10
-t -t
dx=10-t
÷d ÷d
x=(10-t)/d

4 0

3 years ago

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. The area of printed material on the po

grin007 [14]

Answer:

Dimensions of printed poster are

length is 32 cm

width is 48 cm

Step-by-step explanation:

Let's assume

length of printed poster is x cm

width of printed poster is y cm

now, we can find area of printed poster

so, area of printed poster is

$=xy$

we are given that area as 1536

so, we can set it to 1536

$xy=1536$

now, we can solve for y

$y=\frac{1536}{x}$

now, we are given

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm

so, total area of poster is

$A=(8+x+8)\times (12+y+12)$

$A=(x+16)\times (y+24)$

now, we can plug back y

$A=(x+16)\times (\frac{1536}{x}+24)$

now, we have to minimize A

so, we will find derivative

$A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)$

we can use product rule

$A'=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

$=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

now, we can simplify it

$A'=-\frac{24576}{x^2}+24$

now, we can set it to 0

and then we can solve for x

$A'=-\frac{24576}{x^2}+24=0$

$-\frac{24576}{x^2}x^2+24x^2=0\cdot \:x^2$

$-24576+24x^2=0$

$x=32,\:x=-32$

Since, x is dimension

and dimension can never be negative

so, we will only consider positive value

$x=32$

now, we can solve for y

$y=\frac{1536}{32}$

$y=48$

so, dimensions of printed poster are

length is 32 cm

width is 48 cm

5 0

3 years ago

What is the solution to this equation?

slamgirl [31]

Answer: No solution.

Step-by-step explanation:

1. The lines are parallel.

2. If you solve the system of equations you would end up with 0x, so you cannot solve for x.

4 0

3 years ago