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3 years ago

7

The equation d=m/v can be used to calculate the density

1 answer:

ohaa [14]3 years ago

5 0

This is a true statement. An example of using this formula is attached

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Which estimate best describes the area under the curve in square units?

Rufina [12.5K]

My best estimate for this figure is 15

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3 years ago

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Dave earns $150 per week plus 3.8% commission.he sold $4175.68 in the month of February. What is his gross monthly earning for f

Svet_ta [14]

If Dave earns $150/week times 4 weeks = 600

3.8% commission on $4175.68 is $158.68

$600 + $158.68 = $758.68 for the month of February

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4 years ago

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$8.50 yoga mat; 75% markup

Alenkinab [10]

The answer is:
14.875
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3 years ago

Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}

Ede4ka [16]

Step-by-step explanation:

(a)

Using the definition given from the problem

$f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)\\$

Therefore it is true for intersection. Now for union, we have that

$A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]$

Therefore, for this case, it would be true that $f(A\cup B) = f(A)\cup f(B)$ .

(b)

1 is not a set.

(c)

To begin with

$A\cap B \subset A,B$

Therefore

$g(A\cap B) \subset g(A) \cap g(B)$

Now, given an element of $g(A) \cap g(B)$ it will belong to both sets, therefore it also belongs to $g(A\cap B)$ , and you would have that

$g(A)\cap g(B) \subset g(A \cap B)$ , therefore $g(A)\cap g(B) = g(A \cap B)$ .

(d)

To begin with $A,B \subset A \cup B$ , therefore

$g(A) \cup g(b) \subset g(A\cup B)$

7 0

4 years ago

If tanA+cotA=2 then find the value of cosA

aniked [119]

Answer:

cos A=

$\frac{1}{ \sqrt{2} }$

Step-by-step explanation:

We know,

cotA=

$\frac{1}{ \tan(a) }$

Substituting the value of cot A in the given equation, we get

$\tan( a) + \frac{1}{ \tan(a) } = 2$

$\frac{{ \tan }^{2} a + 1}{ \: \tan(a) } = 2$

${ \tan }^{2 \: } a \: + 1 = 2 tan \: a$

${ \tan }^{2} a - 2 \tan \: a + 1 = 0$

$( \tan \: a - 1) {}^{2 } = 0$

$\tan \: a - 1 = \sqrt{0}$

$\tan \: a = 1$

$\tan \: a \: = \tan \: 45$

$a = 45$

$\cos \: a = \cos \: 45$

$\cos \: a = \frac{1}{ \sqrt{2} }$

4 0

3 years ago