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Nuetrik [128]
2 years ago
11

ge\bold\red{{HELP}}" align="absmiddle" class="latex-formula">​

Mathematics
1 answer:
Serggg [28]2 years ago
5 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

A. <u>Understand</u>

The question is asking how much money is saved.

Given :

  • original price = 18795.50

  • lowered price = 15250.95

B. <u>Plan</u>

  • operation to be used here is subtraction

that is :

  • original price - lowered price

C. <u>Solve</u>

  • 18795.50 - 15250.95

  • 3544.55

Hence, savings = Php 3544.55

You might be interested in
2.11 A concert is performed for a crowd of 118
never [62]

Answer:

Number of Students who attended = 22

Number of Adults who attended = 34

Number of Children who attended = 62

Step-by-step explanation:

Total crowd at concert  =  118

Cost of each adult's ticket = $20

Cost of each student's ticket = $16

Cost of each children's ticket = $11.50

The total revenue of the concert = $1745.00

Now, Let the number of students = k

So, the number of children = k + 40

So, number of adults = 118  - {k + (k + 40)}

                                   = 118- 2k -40 = 78 -2k

So, the total revenue = Total revenue from {adults + students + children}

or, $1745.00 = (78 -2k)(20) +  k(16)  +(k+40)(11.50)

here, solving for the value of k :

12.5k = 2020 -  1745

or k = 275\12.5 = 22

So, the Number of students who attended = 22

Number of Adults who attended = 78 - 2k = 78 - 44 = 34

Number of children who attended = k + 40 = 22 + 40 = 62

5 0
3 years ago
How do you solve this limit of a function math problem? ​
hram777 [196]

If you know that

e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving e.

For starters, we have

\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}

Let y=\dfrac34(x+1). Then as x\to\infty, we also have y\to\infty, and

2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2

So in terms of y, the limit is equivalent to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}

Now use some of the properties of limits: the above is the same as

\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}

The first limit is trivial; \dfrac1y\to0, so its value is 1. The second limit comes out to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}

To see why this is the case, replace y=-z, so that z\to-\infty as y\to\infty, and

\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e

Then the limit we're talking about has a value of

\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}

# # #

Another way to do this without knowing the definition of e as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)

(where the notation means \exp(x)=e^x, just to get everything on one line).

Recall that

\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)

if f is continuous at x=c. \exp(x) is continuous everywhere, so we have

\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)

For the remaining limit, write

\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}

Now as x\to\infty, both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83

and our original limit comes out to the same value as before, \exp\left(-\frac83\right)=\boxed{e^{-8/3}}.

3 0
3 years ago
Help that one side is 2x-2
bogdanovich [222]

Answer: x = 2/3

Step-by-step explanation:

<em>Add all sides.</em>

2x-2 + 2x - 2 + x + x =

6x - 4

<em>Put " = 0 "</em>

6x-4 = 0

Add 4 to both sides.

6x-4    =  0

   +4    = +4

-------------------

6x=4

You get 6x = 4 because the 4 on the left side of the equals sign cancelled out!

<em></em>

<em>NOW DIVIDE 6 FROM BOTH SIDES</em>

6x = 4

/6      /6

-----------

x= 4/6

<em>SIMPLIFY / REDUCE FRACTION</em>

4/6 divided by 2 = 2/3

2/3 is the answer !

5 0
3 years ago
What is the first step in describing what figure results when a given plane intersects a given three-dimensional figure
mafiozo [28]
There are three possible results when a plan intersects with a three-dimensional figure and these are:
a point
a line
a plane

The first step in describing the result is to determine first how the plane and the figure intersects.
3 0
3 years ago
What do cos and sin mean?
Mademuasel [1]

Answer and Step-by-step explanation:

These two terms (pronounced as Sine and Cosine) are used to solve for the sides and angles of a triangle in Trigonometry.

They are functions revealing the shape of a right triangle.

Sine is a trigonometric function of an angle. The sine of an acute angle is defined in the context of a right triangle: for the specified angle, it is the ratio of the length of the side that is opposite that angle, to the length of the longest side of the triangle.

Cosine is also a trigonometric function of an angle. The cosine of an angle is the relation of the length of the side that is adjacent that angle, to the length of the longest side of the triangle.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

8 0
3 years ago
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