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3 years ago

How do you solve this type of problem?

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

8 0

The price of ticket last year is $ 1231.65

Solution:

Given that,

This year, Sophia paid $1,071 for season tickets to her favorite baseball team , which is 15% cheaper than last year

To find: Price of season ticket last year

Let the price of season ticket last year be "x"

Therefore, according to question

Price of ticket this year is 15 % cheaper than price of ticket last year

1071 = price of ticket last year - 15 % of 1071

1071 = x - 15 % of 1071

Solve the above expression for "x"

$1071 = x - 15 \% \times 1071\\\\1071 = x - \frac{15}{100} \times 1071\\\\1071 = x - 0.15 \times 1071\\\\1071 = x - 160.65\\\\x = 1071 + 160.65\\\\x = 1231.65$

Thus price of ticket last year is $ 1231.65

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Find P on AB that is 2/3 of the distance from A(-3,-5) to B(9,4).

Ludmilka [50]

A line segment can be divided into ratios.

The coordinate of P on line segment AB is (5,1)

Given

$A = (-3,-5)$

$B = (9,4)$

The position of P from A to B is:

$P_A = \frac 23$

So, from P to B would be:

$P_B =1 - \frac 23$

$P_B = \frac 13$

So, the ratio is:

$m : n = \frac 23 : \frac 13$

Simplify

$m : n = 2 :1$

The coordinate of point P is:

$P = (\frac{mx_2 + nx_1}{m + n},\frac{my_2 + ny_1}{m + n})$

So, we have:

$P = (\frac{2 \times 9 + 1 \times -3}{2 + 1},\frac{2 \times 4 + 1 \times -5}{2 + 1})$

$P = (\frac{15}{3},\frac{3}{3})$

$P = (5,1)$

Hence, the coordinate of P is (5,1)

Read more about line ratios at:

brainly.com/question/8847082

8 0

2 years ago

Please help me :) with this math problem, it's exponential functions.

noname [10]

Answer:

f(x) = 9(5/3)^x

Step-by-step explanation:

f(x) = a(b^x)

x = 0

9 = a(b⁰)

9 = a

f(x) = 9(b^x)

x = 1

15 = 9(b¹l

b = 15/9

b = 5/3

f(x) = 9(5/3)^x

8 0

3 years ago

Given:• PQRS is a rectangle.• mZ1 = 50°Р21SRWhat is mZ2?130°85°70°65°

Over [174]

To answer this question, we need to recall that: "the diagonals of a rectangle bisect each other"

Thus, if we assign the point of intersection of the two diagonals in the rectangle as point O, we can say that the triangle OQR is an "isosceles triangle". Note that this is because the lengths OR and OQ are equal since we know that: "the diagonals of a rectangle bisect each other". See the below diagram for clarity.

Now, we have to recall that:

- the base angles of any isosceles triangle are equal. This is a fact, and this means that the angles

- also the sum of all the angles in any triangle is 180 degrees

Now, considering the isosceles triangle OQR, we have that:

$\angle OQR+\angle ORQ+\angle ROQ=180^o$

Now, since the figure already shows that angle $m\angle2+\angle ORQ+50^o=180^o$ Now, since we have established that the base angles $m\angle2+m\angle2+50^o=180^o$ we can now solve the above equation for m<2 as follows:

$\begin{gathered} m\angle2+m\angle2+50^o=180^o \\ \Rightarrow2m\angle2+50^o=180^o \\ \Rightarrow2m\angle2=180^o-50^o \\ \Rightarrow2m\angle2=130^o \\ \Rightarrow m\angle2=\frac{130^o}{2}=65^o \end{gathered}$

Therefore, the correct answer is: option D