6,237.2<br> The 2 in hundreds place is___ the value of the 2 in the tenths place.

At 6:00 a.m., the outside temperature was –14.4⁰. By noon, the outside temperature had dropped 8.1⁰. What was the outside temper

Advocard [28]

-14.4 - 8.1 = - 22.5

The outside temperature at noon is - 22.5 degrees

5 0

2 years ago

Read 2 more answers

In an axiomatic system, which category do points, lines, and planes belong to?

scZoUnD [109]

Undefined terms is the category its belongs to.

3 0

2 years ago

Patrice goes to fishers candy shop with the price of gummy candies modeled by the equation y=5/2x where x is the amount of gummy

icang [17]

Answer:

You need to put the question, but I'll try to help as much as I can without the question. y=5/2x, if you put the amount of gummy candy in lbs in the x's place, and you multiply the amount by 5/2, you will get the price.

Step-by-step explanation:

7 0

2 years ago

A transformer increases the input voltage from 12V to 12,000V. What is the ratio of the number of wire turns on the primary coil

erma4kov [3.2K]

Answer:

$\frac{n_p}{n_s}=\frac{1}{1000}$

Step-by-step explanation:

Given voltage of primary coil is $V_p=12V$

And voltage of secondary coil is $V_s=12,000V$

A transformer is a device that contains primary and secondary coil. These coils are mounted on soft iron. The change in voltage depends upon the number of turns of primary coil $(n_p)$ and secondary coil $(n_s)$ .

Following equation represents the relation between them.

$\frac{V_s}{V_p} =\frac{n_s}{n_p}\\\\\frac{12,000}{12}=\frac{n_s}{n_p}\\\\1000=\frac{n_s}{n_p}$

Then, ratio of the number of wire turns on the primary coil to the number of turns on the secondary coil is

$\frac{n_p}{n_s}=\frac{1}{1000}$

5 0

2 years ago

Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$

6 0

2 years ago

6,237.2 The 2 in hundreds place is___ the value of the 2 in the tenths place.