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3 years ago

PICTURE BELOW / What is the volume of this cube?

Mathematics

2 answers:

Nata [24]3 years ago

6 0

The answer should be 74.08

Rama09 [41]3 years ago

3 0

Answer:

74.088

or 74.09

Step-by-step explanation:

4.2

x4.2

74.088

<em>Rounded</em>

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there are 7 chimpanzees at the zoo the mother chimpanzee usually has a baby every 4 years how many babies can she have in 28 yea

Finger [1]

Are you saying that every chimpanzee is a mother? If so, then each chimpanzee would have a baby 7 times in 28 years, meaning there would be 49 chimpanzees.

If you mean just one, then only 7 babies in 28 years.

4 0

3 years ago

What is the mean of 2,7,5,4,3,3,5,5,6,5

kkurt [141]

Answer:

4.5

Step-by-step explanation:

To solve for the mean (average), you need to add together all the numbers then divide by how many numbers there are. In this case, there are 10 numbers.

2+7+5+4+3+3+5+5+6+5 = 45

45/10 = 4.5

3 0

3 years ago

Read 2 more answers

The perimeter of a rectangle olis 40 inches. If the length of the rectangle is 13 inches, which equation could be used to find t

Elis [28]

The perimeter of a rectangle olis 40 inches. If the length of the rectangle is 13 inches, which equation could be used to find the width of x

The width is 7

5 0

4 years ago

1/4+1/10 Add or subtract. Write the answer in simplest form.

vagabundo [1.1K]

Answer:

7/20

Step-by-step explanation:

To add fractions, the first step is to get a common denominator

1/4 + 1/10

The common denominator is 20

1/4 *5/5 = 5/20

1/10 *2/2 = 2/20

5/20 + 2/20 = 7/20

5 0

2 years ago

Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A

mash [69]

Answer:

Step-by-step explanation:

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA$

we take the LHS so here goes,

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\$

since , $sin^2A+cos^2A=1$

the identity becomes,

$(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\$

now, we know,

$sin^2A=1-cos^2A$ and $cos^2A=1-sin^2A$

the identity becomes,

$(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\$

$(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})$

sin A and cos A cancel out it becomes zero

$\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\$

Splitting the denominator the identity becomes

$\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A} \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\$

Hence,

$tanAsecA-cotAcosecA$

3 0

3 years ago