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sdas [7]
3 years ago
15

Find the volume of the figure below. 5 cm 11 cm 66 cm 82.5 cm 132 cm 165 cm​

Mathematics
1 answer:
stealth61 [152]3 years ago
5 0

Answer:

A) 66cm³

Step-by-step explanation:

Using the Pythagorean Theorem, the base of the triangle is 4.  The area of a triangle is 1/2 base times height.  Then you multiply times 11 to get the volume.

1/2(3)(4)(11)=66

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A deck of 52 cards is shuffled. The first five cards of the deck are red. What is the probability that the last card in the deck
saveliy_v [14]

Answer:

The answer is "\frac{1}{47}".

Step-by-step explanation:

Although the five first players are red, the very first five cards are not spaded as the spad is a black card.  

Its chance of the final card mostly on deck being the ace of spades is therefore determined as:

= \frac{1}{\text{Number of remaining cards with the initial 5 cards}} \\\\= \frac{1}{47}

Thus, \frac{1}{47} here is the necessary chance.

7 0
3 years ago
17 mm<br> 8 mm<br> 9 mm<br> 15 mm
OleMash [197]

Answer:

do you have to add the numbers together or multiply

4 0
3 years ago
Read 2 more answers
The volume of the crate is 72 cubic feet but the length is 3ft the base is 6ft and the height is unknown
k0ka [10]
The volume of a crate is b x l x h. Reverse it to get 3 times 6 times h = 72. 3 times 6 is 18, so you know 18h=72. Divide both sides by 18 to get h = 4 ft.
3 0
3 years ago
A 15-meter by 23-meter garden is divided into two sections. Two sidewalks run along the diagonal of the square section and along
Alisiya [41]

the picture in the attached figure


Applying the Pythagorean Theorem

Find the lengths of the two sidewalks


Step 1

square section

D=√{15²+15²}

D=√450 m

D=21.21 m


Step 2

rectangular section

D=√{15²+8²}

D=√289 m

D=17 m


Step 3

the sum of the lengths of the diagonals of the two sections is

17m + 21.21 m = 38.21 m


the answer is

38.2 m


6 0
3 years ago
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Simplify the following expression. cot^2x secx-cosx
Solnce55 [7]

ANSWER

\cos(x)   \cot ^{2} (x)

EXPLANATION

The given expression is;

\cot^{2} (x)  \sec(x)  -  \cos(x)

Change everything to

\sin(x)

and

\cos(x)

This implies that,

\frac{ \cos^{2} (x) }{ \sin^{2} (x) }  \times ( \frac{1}{ \cos(x) } )-  \cos(x)

Cancel the common factors,

\frac{ \cos(x) }{ \sin^{2} (x) }  \times ( \frac{1}{1} )-  \cos(x)

\frac{ \cos(x) }{ \sin^{2} (x) }-  \cos(x)

\frac{ \cos(x)  -  \sin ^{2} (x)  \cos(x) }{ \sin^{2} (x) }

= \frac{ \cos(x)(1  -  \sin ^{2} (x) ) }{ \sin^{2} (x) }

= \frac{ \cos(x)(\cos^{2} (x) ) }{ \sin^{2} (x) }

= \cos(x)  \cot^{2} (x)

6 0
3 years ago
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