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djverab [1.8K]
3 years ago
5

Find the area of a regular hexagon with an apothem of 15 cm. DO

Mathematics
1 answer:
Ksivusya [100]3 years ago
6 0

Answer: To Find the area you would need to use Area= 1/2 a multiplied by the perimeter. if you do not have the perimeter then you can find it by multiplying # of sides and length of sides. Hope this helps. : )

Step-by-step explanation:

a=apothem

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On a shopping trip kellen spent $3 on a bus ticket $9 on a book and $5 after the trip . How much money did he start with
BaLLatris [955]

He would have $17 9+3=12 12+5=17


4 0
3 years ago
A function is shown: f(x) = 36x² - 1.
givi [52]

Based on the given function, the equivalent function that best shows the x-intercepts on the graph is f(x) = (6x - 1)(6x + 1)

<h3>What are equivalent functions?</h3>

Equivalent functions are different functions that have equal values when evaluated and compared

<h3>How to determin the equivalent function that best shows the x-intercepts on the graph?</h3>

The function is given as:

f(x) = 36x^2 - 1

Express 1 as 1^2

f(x) = 36x^2 - 1^2

Express 36x^2 as (6x)^2

f(x) = (6x)^2 - 1^2

Apply the difference of two squares.

This is represented as:

(a + b)(a - b) = a^2 - b^2

So, we have the following equation

f(x) = (6x - 1)(6x + 1)

Based on the given function, the equivalent function that best shows the x-intercepts on the graph is f(x) = (6x - 1)(6x + 1)

Read more about equivalent function at:

brainly.com/question/2972832

#SPJ1

8 0
2 years ago
B to the eighth power multiplied by b to the fourth power
scZoUnD [109]
U just add the exponents.

b^8 x b^4 = b^12

hope this helps :)





7 0
3 years ago
There are 6 roses in a vase of 11 flowers. The rest are daisies. What is the ratio of all flowers in the vase to daisies (PICTUR
lara [203]

Answer:

a) 6:5

b) 11:5

(Daisies: 11-6=5)

5 0
2 years ago
Question 7 of 10
Ganezh [65]

Answer:

  • Option B

Step-by-step explanation:

Given Equation :

\qquad \sf \dashrightarrow \: 3(4x+3) = 2x - 5(3 - x) + 2

Using distribute property:

\qquad \sf \dashrightarrow \: 12x + 9 = 2x - 15 + 5x + 2

Adding the like terms we get :

\qquad \sf \dashrightarrow \: 12x + 9 = 2x  + 5x  - 15 + 2

\qquad \sf \dashrightarrow \: 12x + 9 = 7x  - 13

Transposing the variables on the right side and constant terms on the left side :

\qquad \sf \dashrightarrow \: 12x  - 7x =   - 13 - 9

\qquad \sf \dashrightarrow \: 5x =   - 22

Dividing both sides by 5 :

\qquad \sf \dashrightarrow \:  \dfrac{5x}{5}  =  \dfrac{ - 22}{5}

\qquad \bf \dashrightarrow \:  x =  \dfrac{ - 22}{5}

3 0
2 years ago
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