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3 years ago

Simplify fully: (Question is on the photo).

Mathematics

2 answers:

makkiz [27]3 years ago

8 0

Answe

8x+9

Step-by-step explanat

8(x+6x+9)/2x+6

4(x+x+9)

8x+9

dedylja [7]3 years ago

6 0

Step-by-step explanation:

$4(x + 3) {z}^{ - 1}$

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A ticket to a particular theme park regularly costs $84.99, however the theme park is running a spring break special for 15% off

Ilya [14]

Answer: 306.30

Step-by-step explanation:

DiscountPrice = 84.99 - (84.99 x 0.15)

Cost for 4 = discountPrice x 4

Total = cost for 4 x 1.06

5 0

2 years ago

Read 2 more answers

A walkway goes through the middle of a square playground from one corner to the opposite corner. If the walkway is 20 meters lon

Dahasolnce [82]

I think you would use the Pythagorean Theorem to solve this, as a square cut across diagonally creates two isocele triangles. Since the longest side is 20 m, this value would be imputed into c^2.

a^2 + b^2 = c^2
a^2 + a^2 = 20^2
2a^2 = 400
2a^2/2 = 400/2
a^2 = 200
a = 14.14

Thus, each sides of the playground are 14.14 meters long.

7 0

2 years ago

If g(x)=f(1/3x) which statement is true

uysha [10]

Answer:

the graph of g(x) is horizontally stretched by a factor of 3

Step-by-step explanation:

8 0

3 years ago

Situation:

Gala2k [10]

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

Given formula:

$\sf N=N_0e^{-kt}$

$\sf N_0$ = initial mass (at time t=0)
N = mass (at time t)
k = a positive constant
t = time (in days)

Given values:

$\sf N_0$ = 11 g
k = 0.125

Half-life: The time required for a quantity to reduce to half of its initial value.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

$\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}$