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2 years ago

Find the area of a regular pentagon with an apothem of 2.75 cm and a side length of 4 cm. (round to the nearest whole number)

Mathematics

1 answer:

valentinak56 [21]2 years ago

4 0

Answer:B 28 cm^2

Step-by-step explanation:

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What must be added to the expression below to make it a perfect square trinomial? *

artcher [175]

I think the answer is 24

7 0

3 years ago

72 is what percent of 45? __ = p/100 p= __

Marta_Voda [28]

Answer:

72 of 45 is 160%

Step-by-step explanation:

72 of 45 can be written as:

$\frac{72}{45}$

To find percentage, we need to find an equivalent fraction with denominator 100. Multiply both numerator & denominator by 100

$\frac{72}{45} *\frac{100}{100}$

$=(\frac{72*100}{45} )*\frac{1}{100} =\frac{160}{100}$

Therefore, the answer is 160%

If you are using a calculator, simply enter 72÷45×100 which will give you 160 as the answer.

3 0

2 years ago

Read 2 more answers

Ava has two frogs. This is 1/4 the number of frogs that Heather has. How many frogs does Heather have?

oksano4ka [1.4K]

Answer:

Step-by-step explanation:

2 x 4

7 0

3 years ago

Read 2 more answers

A bag contains 5 2/5kg of sand. The bag has a hole in it. Every 10 minutes, 1 2/3kg of sand escapes from the bag. After 20 minut

fenix001 [56]

Quantify of sand in a bag =

$= \: 5 \frac{2}{5} \ \:$

$= \frac{(5 \times 5 +2) }{5}$

$= \frac{27}{5} \: kg$

Quantity of sand that escapes in every 10 minutes :-

$= 1 \frac{2}{3} kg$

Quantity of sand that will escape in 20 minutes :-

$= 1 \frac{2}{3} \times 2$

$= \frac{(3 \times 1 + 2)}{3} \times 2$

$= \frac{5}{3} \times 2$

$= \frac{10}{3}$

Quantity of sand that will remain in the sand bag :-

$= \frac{27}{5} - \frac{10}{3}$

$= \frac{(27 \times 3)}{15} - \frac{(10 \times 5)}{15}$

$= \frac{81}{15} - \frac{50}{15}$

$= \frac{31}{15}$

$= 2 \frac{1}{5} \: kg$

$therefore \: , \: 2 \frac{1}{5} \: kg \: sand \: is \: left \: in \: the \: bag \: .$

3 0

3 years ago

PLEASE HELPP MEE ASAPP!!

WARRIOR [948]

Answer:

Step-by-step explanation:

each zero of the function will have a factor of (x - x₀)

h(x) = a(x + 3)(x + 2)(x - 1)

h(x) = a(x + 3)(x² + x - 2)

h(x) = a(x³ + 4x² + x - 6)

or the third option works if a = 1

however this equation gives us the points (0, -6) and (-1. -4), so "a" must be -2

h(x) = -2x³ - 8x² - 2x + 12

to fit ALL of the given points as it fits the three zeros and also h(0) and h(-1) so I guess that is why the given group is a partial set of solution sets

7 0

2 years ago