All categories

3 years ago

Which statement belongs in the intersection of the Venn Diagram? (4 points)

Mathematics

2 answers:

alina1380 [7]3 years ago

8 0

"Measured using the distance formula" is the one statement among the following choices given in the question that <span>belongs in the intersection of the Venn Diagram. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope the answer helps you.</span>

Over [174]3 years ago

3 0

Answer:

<h3><u>Measured using the distance formula</u></h3>

Step-by-step explanation:

Took the test and got it correct

You might be interested in

I GIVE BRAINLIEST PLS HELP

ioda

The answer is 8
Here's why:
${ ( \frac{( {6}^{7}) \times ( {3}^{3}) }{( {6}^{6}) \times ( {3}^{4} ) } )}^{3} = \\ ( \frac{6}{3} ) ^{3} = \\ \frac{216}{27} = 8$
The exponents are subtracted one from another when divided.
$\frac{ a ^{b} }{ {a}^{c} } = {a}^{b - c}$
We can look at the problem this way:
$( \frac{6^{7} }{6 ^{6} } \times \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6} \times {3}^{3 - 4} ) = \\ ({6}^{1} \times {3}^{ - 1} )$
Since we have the power of -1 on the 3, we apply this rule:
${a}^{ - b} = \frac{1}{ {a}^{b} }$
Also this rule because we have the power of 1 on the 6:
${a}^{1} = a$
Then we get this:
$(6 \times \frac{1}{3} )^{3} = ( \frac{6}{3} )^{3}$
We apply the rule:
$( \frac{a}{b}) ^{c} = \frac{ {a}^{c} }{ {b}^{c} }$
We get this:
$\frac{{6}^{3} }{ {3}^{3} } = \frac{216}{27} = 8$

4 0

3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$