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2 years ago

How do I find the area of an octagon which has 7m on each side and the height is 8.8m

Mathematics

1 answer:

Naddika [18.5K]2 years ago

5 0

Answer:

236.59

Step-by-step explanation:

The formula 2(1+ $\sqrt{2}$ ) $l^{2}$ can be used. Substitute the length of one side into the equation and solve.

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Find the volume of the cone

SVEN [57.7K]

Answer:

gkali 53 *622 kalia qqua 5% habwa

3 0

2 years ago

F(x) = 3/(x + 2) - sqrt(x - 3) complete the following statement: f(19)=

liberstina [14]

Answer:

f(19)=3/21-4=-3.857

Step-by-step explanation:

f(x)=3/(x+2)-sqrt(x-3)

f(19)=3/(19+2)-sqrt(19-3)

f(19)=3/21-sqrt(16)

f(19)=3/21-4

f(19)=-3.857

5 0

3 years ago

Dave is comparing the circumferences of several trees in his yard. The oak tree is 0.539 meters in circumference, the ash tree h

stepan [7]

Answer: c. elm

Step-by-step explanation:

Given : The circumference of the oak tree= 0.539 meters

= $0.539\times100=53.9\text{ cm}$ [∵ 1 m = 100 cm] (1)

The circumference of the ash tree= 0.509 yards

$=0.509\times3\text{ feet}$ [∵ 1 yard = 3 feet]

$=0.509\times3\times 30.48\text{ cm}$ [∵ 1 foot = 30.48 cm]

$=46.54296\approx46.54\text{ cm}$ (2)

The circumference of the elm tree = 6281.70 millimeters

$=\dfrac{6281.70}{10}=628.17\text{ cm}$ (3)

The circumference of the poplar tree = 0.000385 miles

$=0.000385 \times5280\text{ feet}$ [∵ 1 mile = 5280 feet]

$=0.000385 \times5280\times30.48\text{ feet}$ [∵ 1 foot = 30.48 cm]

$=61.959744\approx61.96\ \text{ cm}$ (4)

From (1) , (2) , (3 ) and (4) it is clear that

46.54< 53.9 < 61.96 < 628.17

Hence, the elm tree has the greatest circumference.

5 0

3 years ago

Evaluate: 6-(2/3)^2 A. 17/3 B. 52/3 C. 49/9 D. 50/9

Komok [63]

Answer:

The correct answer is: "Option [D]".

Step-by-step explanation:

Hi student, let me help you out!

....................................................................................................................................

Let's use the acronym PEMDAS. With the help of this little acronym, we will not make mistakes in the Order of Operations! :)

$\dag\textsf{Acronym \: PEMDAS}$

P=Parentheses,

E=Exponents,

M=Multiplication,

D=Division,

A=Addition,

S=Subtraction.

Now let's start evaluating our expression, which is $\mathsf{6-(\cfrac{2}{3})^2}$

According to PEMDAS, the operation that we should perform is "E-Exponents".

Notice that we have a fraction raised to a power. When this happens, we raise both the numerator (2 in this case) and the denominator (3 in this case) to that power, which is 2. After this we obtain $\mathsf{6-\cfrac{4}{9}}$ .

See, we raised both the numerator and the denominator to the power of 2.

Now what we should do is subtract fractions.

Note that 6 and -4/9 have unlike denominators. First, let's write 6 as a fraction: $\mathrm{\cfrac{6}{1}-\cfrac{4}{9}}$ . Now let's multiply the denominator and the numerator of the first fraction times 9: $\mathrm{\cfrac{54}{9}-\cfrac{4}{9}}$ .

See, now the fractions have the same denominator. All we should do now is subtract the numerators: $\mathrm{\cfrac{50}{9}}$ .