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3 years ago

What is 2 divided by 19, and the steps to solving it? PLEASE HELP ME!!!!!

Mathematics

1 answer:

Serga [27]3 years ago

6 0

19 divided by 2 is 9.5 Its a decimal because no whole numbers fit into it.

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42 less than the product of 28 and an unknown number is 154

zalisa [80]

First you write out your equation, which is 28x-42=154. Then you solve you equation. First, add 42 to each side, get 28x=196. Divide both sides by 28, get x=7.

8 0

3 years ago

This is a substitution problem

mrs_skeptik [129]

They consists of equations and unknown variables. There are 4 PCs and 8 Mac in the room

<h3>System of equation</h3>

They consists of equations and unknown variables

Let the number of PCs be x

Let the number of Mac be y

If Mke has 12 computers, then;

x + y = 12

If there are two Macs for every PC in his classroom, then;

y = 2x

Substitute

x + 2x = 12

3x = 12

x = 4

Recall y = 2x

y = 2(4)

y = 8

There are 4 PCs and 8 Mac in the room

Learn more on system of equations here; brainly.com/question/14323743

#SPJ2

6 0

2 years ago

Read 2 more answers

I need help, can someone please explain this step by step.

Helga [31]

180-117.3=62.7

answer is 62.7

3 0

2 years ago

Any of these please I’m having difficulty

frosja888 [35]

Answer:

864 squared meters

Step-by-step explanation:

Set up a proportion. The scale between 270 and 360 should be the same as 648 to the fourth yard.

$\frac{270}{360} = \frac{648}{x}\\270x = (360)(648)\\270x = 233280\\x = 864 m^2$

7 0

2 years ago

A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 3 feet, 55 feet awa

tatuchka [14]

Answer:

Second puck travels farther

Step-by-step explanation:

Maximum height of first puck = 3 feet

The height of a second hockey puck is modeled by:

$y=x\left(0.15-0.001x\right)$

$y=0.15x-0.001x^2$

To find maximum height of second puck

$y'=0.15-0.002x$

Equate the derivative equals to 0

0.15-0.002x=0

$\frac{0.15}{0.002}=x$

75 = x

At x = 75

$y=0.15(75)-0.001(75)^2=5.625$

So, The maximum height of second puck is greater than first puck

So, Second puck travels farther

5 0

2 years ago