What is<br><br>(\frac{-7c}{p3})^-4[/tex] <br><br>without negative exponents

Kool-aide is made from flavored powder and water, with the ratio of powder to water 3:2 based on weight. If 120g of flavored pow

MAXImum [283]

The given ratio of powder to water is 3:2 based on weight for Kool-aide. It is also given that 120g of flavored powder is needed for a certain amount of Kool-aide.

Let us assume that the amount of water (by weight) required to maintain the given ratio of 3:2 be represented by the letter $x$ . Thus, from the given information we can conclude that:

$\frac{Powder Weight}{Water Weight}=\frac{3}{2}=\frac{120}{x}$

$\therefore \frac{3}{2}=\frac{120}{x}$

Cross-multiplying we get:

$3x=240$

$\therefore x=\frac{240}{3}=80$ grams.

Thus, the amount of water in grams in that Kool-aide is 80 grams.

5 0

2 years ago

Read 2 more answers

Indicate the equation of the given line in standard form. Show all your work for full credit. the line containing the median of

alukav5142 [94]

Answer:

* The equation of the median of the trapezoid is 10x + 6y = 39

Step-by-step explanation:

* Lets explain how to solve the problem

- The slope of the line whose end points are (x1 , y1) , (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid point of the line whose end point are (x1 , y1) , (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

- The standard form of the linear equation is Ax + BC = C, where

A , B , C are integers and A , B ≠ 0

- The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

- It has two properties:

# It is parallel to both bases

# Its length equals half the sum of the base lengths

* Lets solve the problem

- The trapezoid has vertices R (-1 , 5) , S (! , 8) , T (7 , -2) , U (2 , 0)

- Lets find the slope of the 4 sides two find which of them are the

parallel bases and which of them are the non-parallel bases

# The side RS

∵ $m_{RS}=\frac{8-5}{1 - (-1)}=\frac{3}{2}$

# The side ST

∵ $m_{ST}=\frac{-2-8}{7-1}=\frac{-10}{6}=\frac{-5}{3}$

# The side TU

∵ $m_{TU}=\frac{0-(-2)}{2-7}=\frac{2}{-5}=\frac{-2}{5}$

# The side UR

∵ $m_{UR}=\frac{5-0}{-1-2}=\frac{5}{-3}=\frac{-5}{3}$

∵ The slope of ST = the slop UR

∴ ST// UR

∴ The parallel bases are ST and UR

∴ The nonparallel sides are RS and TU

- Lets find the midpoint of RS and TU to find the equation of the

median of the trapezoid

∵ The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

∵ The midpoint of RS = $(\frac{-1+1}{2},\frac{5+8}{2})=(0,\frac{13}{2})$

∵ The median is parallel to both bases

∴ The slope of the median equal the slopes of the parallel bases = -5/3

∵ The form of the equation of a line is y = mx + c

∴ The equation of the median is y = -5/3 x + c

- To find c substitute x , y in the equation by the coordinates of the

midpoint of RS

∵ The mid point of Rs is (0 , 13/2)

∴ 13/2 = -5/3 (0) + c

∴ 13/2 = c

∴ The equation of the median is y = -5/3 x + 13/2

- Multiply the two sides by 6 to cancel the denominator

∴ The equation of the median is 6y = -10x + 39

- Add 10x to both sides

∴ The equation of the median is 10x + 6y = 39

* The equation of the median of the trapezoid is 10x + 6y = 39

7 0

3 years ago

A submarine sits at -300 meters in relation to sea level. Then it descends 115 meters. What is it’s new position in relation to

Anna71 [15]

Answer:

A. -415m

Step-by-step explanation:

-300 meters is the starting point.

Then, the submarine goes 115m further below the surface, which can be represented as -115.

Add -300 and -115 and you get -415m.

3 0

3 years ago

Question 3.3. Find the LCD of these fractions.

Yakvenalex [24]

Did you ever get an answer?

7 0

3 years ago

-6-5[-4-(6+12)]+(-5)

Luden [163]

Answer:

99

Step-by-step explanation:

-6-5[-4-(6+12)]+(-5)

Following PEMDAS

We work the parentheses from the inside out

-6-5[-4-(18)]+(-5)

-6-5[-22]+(-5)

Now we multiply

-6+110 -5

Now we add and subtract from left to right

104 -5

99

6 0

3 years ago

What is(\frac{-7c}{p3})^-4[/tex] without negative exponents