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3 years ago

What is 5/2 times 24/1? (explanation please!)

Mathematics

1 answer:

Marianna [84]3 years ago

5 0

If you mean multiply this is it;

5/2 * 24/1

5 * 24 / 2 * 1

120 / 2

60.
hope this helps, God bless!

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We know the following about the numbers a, b and c:

labwork [276]

Step-by-step explanation:

(a + b)² = 9

(b + c)² = 25

(a + c)² = 81

Taking the square root:

a + b = ±3

b + c = ±5

a + c = ±9

By adding these three equations together and dividing both sides by 2, we get the value of a + b + c.

Possible combinations for a + b + c such that the sum is greater than or equal to 1 are:

a + b + c = (-3 + 5 + 9)/2 = 11/2

a + b + c = (3 − 5 + 9)/2 = 7/2

a + b + c = (3 + 5 + 9)/2 = 17/2

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3 years ago

what is 7 divided by 1/3 in multiplication is it A.1/3 multiplied by 7 =k B. 1/3 multiplied by k=7 or C. 1/3=k multiplied by 7.

NeX [460]

Answer:

B. 1/3 multiplied by k=7

Step-by-step explanation:

7 ÷ ⅓ = k

Multiply both sides by ⅓

7 ÷ ⅓ × ⅓ = k × ⅓

7 = k × ⅓

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3 years ago

I need help please and explain me thanks

erastova [34]

Day 1 and Day 3 his score add up to zero.
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3 years ago

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Does the equation 2x+ 3y=-6, 4x+6y=-12 have infinetley many solutions

Vadim26 [7]

Answer:

Yes it has infinity of solutions.

Step-by-step explanation:

If you plug in certain numbers it will make the statements true.

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2 years ago

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Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ

-Dominant- [34]

Answer:

$\tan(\beta)$

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something. If you have options that you're building toward, aim toward one of them.

${\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}$ and ${\sec(\theta)}={\dfrac{1}{\cos(\theta)}$

Recall the following reciprocal identity:

$\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}$

So, the original expression can be written in terms of only sines and cosines:

$\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)$

$\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }$

$\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}$

$\sin(\beta) *\dfrac{1 }{\cos(\beta) }$

$\dfrac{\sin(\beta)}{\cos(\beta) }$

Working toward one of the answers provided, this is the tangent function.

The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero. However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to $\tan(\beta)$ .

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