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3 years ago

Log(100) = log(400) - log(4) "

1 answer:

5 0

When working with logarithms, remember that a subtraction sign means you can divide.
log(400) - log(4) can be turned into log(400/4), which is further simplified to log(100).

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There are 75 students enrolled in a camp. The day before the camp begins, 8% of the students cancel. How many students actually

8090 [49]

Answer:

69 students attend the camp

Step-by-step explanation:

8% of 75 = 6

75 - 6 = 69

6 0

3 years ago

Read 2 more answers

123333333333333333333

masya89 [10]

Because of the area

7 0

3 years ago

. Given ????(5, −4) and T(−8,12):

damaskus [11]

Answer:

a) $y=\dfrac{13x}{16}-\dfrac{129}{16}$

b) $y = \dfrac{13x}{16}+ \dfrac{37}{2}$

Step-by-step explanation:

Given two points: $S(5,-4)$ and $T(-8,12)$

Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!

Perpendicular to ST:

the equation of any line is given by: $y = mx + c$ where, m is the slope(also known as gradient), and c is the y-intercept.

to find the perpendicular of ST <u>we first need to find the gradient of ST, using the gradient formula.</u>

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)

$m = \dfrac{12 - (-4)}{(-8) - 5}$

$m = \dfrac{-16}{13}$

to find the perpendicular of this gradient: we'll use:

$m_1m_2=-1$

both $m_1$ and $m_2$ denote slopes that are perpendicular to each other. So if $m_1 = \dfrac{12 - (-4)}{(-8) - 5}$ , then we can solve for $m_2$ for the slop of ther perpendicular!

$\left(\dfrac{-16}{13}\right)m_2=-1$

$m_2=\dfrac{13}{16}$ :: this is the slope of the perpendicular

a) Line through S and Perpendicular to ST

to find any equation of the line all we need is the slope $m$ and the points $(x,y)$ . And plug into the equation: $(y - y_1) = m(x-x_1)$

side note: you can also use the $y = mx + c$ to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

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

we have the slope of the perpendicular to ST i.e $m=\dfrac{13}{16}$

and the line should pass throught S as well, i.e $(5,-4)$ . Plugging all these values in the equation we'll get.

$(y - (-4)) = \dfrac{13}{16}(x-5)$

$y +4 = \dfrac{13x}{16}-\dfrac{65}{16}$

$y = \dfrac{13x}{16}-\dfrac{65}{16}-4$

$y=\dfrac{13x}{16}-\dfrac{129}{16}$

this is the equation of the line that is perpendicular to ST and passes through S

a) Line through T and Perpendicular to ST

we'll do the same thing for $T(-8,12)$

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

$(y -12) = \dfrac{13}{16}(x+8)$

$y = \dfrac{13x}{16}+ \dfrac{104}{16}+12$

$y = \dfrac{13x}{16}+ \dfrac{37}{2}$

this is the equation of the line that is perpendicular to ST and passes through T

7 0

3 years ago

A circle passes through point (-2, -1) and its center is at (2, -1). Which equation represents the circle?

Anastaziya [24]

Answer:

$\text{A)}\qquad (x-2)^2+(y+1)^2=16$

Step-by-step explanation:

The formula for a circle of radius r centered at (h, k) is ...

(x -h)^2 +(y -k)^2 = r^2

Both of the given points are on the line y=-1. The distance between them is the difference of their x-coordinates, 2 -(-2) = 4. So, the radius of the circle is 4 and the equation becomes ...

(x -2)^2 +(y -(-1))^2 = 4^2

(x -2)^2 +(y +1)^2 = 16 . . . . . . . . . matches choice A

7 0

3 years ago

Evaluate 32 + 12 = ( 6 – 3) x 8.

Brrunno [24]

Hello there!

(Equation #1) 32 + 12 = 44

(Equation #2) 6 - 3 = 3, 3 x 8 = 24

These both equations don't have the same answer, even though it says that they are equal with each other, if you could correct these these 2 equations I am willing to answer again.

Hope this helped!!

6 0

3 years ago

Read 2 more answers