Solve the following using 2 partial product. 73×23

Please help me out. I don’t understand this problem.. An equation of a line through (-7,-10) which is perpendicular to the line

nordsb [41]

Answer:

The Equation:

$y = \frac{1}{3}x - \frac{23}{3}$

The y-intercept:

$-\frac{23}{3}$

Step-by-step explanation:

Things you need to know:

Perpendicular mean opposite slope.

Need to know point slope form of a linear equation $y - y_1 = x(x - x_1)$

Ok, with that out the way, lets get started!

From the equation y = -3x + 2 we can see the slope, which is -3. We are asked to find an equation that is perpendicular to that slope so that means we need an opposite slope. To turn this slope to an opposite slope flip it and change the sign

-3 = $\frac{-3}{1} = \frac{1}{3}$

Now that we have our opposite slope, which is $\frac{1}{3}$ , we can find the equation the problem is asking us to solve.

We do this by using the point (-7, -10) and our new slope, which is $\frac{1}{3}$ and by using $y - y_1 = x(x - x_1)$

Insert our point and slope into $y - y_1 = x(x - x_1)$ and solve for y

$y - y_1 = x(x - x_1)$

$y - (-10) = \frac{1}{3}(x - (-7))$

$y + 10) = \frac{1}{3}(x + 7))$

Work on the right side of the = sign

Distribute 1/3 to x and 7

$y + 10 = \frac{1}{3}(x + 7))$

$y + 10 = \frac{1}{3}x + \frac{7}{3}$

Bring the 10 over to the right side by adding a -10

$y + 10 -10= \frac{1}{3}x + \frac{7}{3} - 10$

$y = \frac{1}{3}x + \frac{7}{3} - 10$

Do the math for $\frac{7}{3} - 10 = -\frac{23}{3}$

$y = \frac{1}{3}x - \frac{23}{3}$

We are done.

The Equation:

$y = \frac{1}{3}x - \frac{23}{3}$

The y-intercept:

$-\frac{23}{3}$

7 0

3 years ago

Can someone help and explain ? thank you !

AfilCa [17]

$\sf Hello!$

$\sf y = 2x^{2} + 4x - 3$

$\sf Discriminant$ :

= $\sf b^{2} - 4ac$

= $\sf (4)^{2} - 4.(2).(-3)$

= $\sf 16 - (-24)$

= $\sf 16 + 24$ = $\sf 40$

$\sf Since,$
$\sf D \:is\:Greater\: than\: 0.$

$\sf Roots\: are\: Real\: and\: Distinct$

$\sf Hence,$

$\sf y = 2x^{2} + 4x - 3\: have\: A.\: Two\: real\: roots$

~ $\sf iCarl$

7 0

4 years ago

when rounded to the nearest hundred the number of pupils in a school is 1800 what are the smallest and largest numbers of pupils

barxatty [35]

1749 should be rounded to 1700 because 1749 is closer to 1700 than to 1800

1750 should be rounded to 1800 because it is equally close to 1700 and 1800, and so you apply the rule than when the number to round finish in 5 you round to the next upper number.

Now, if you apply the same 1849 should be tha greater number that when rounded to hundreds becomes 1800.

So, the smalles and largest numbers of pupils there could be in the school are 1849 and 1750, respectively.

5 0

4 years ago

Please show your work

Allushta [10]

Answer:

$x = 56$