All categories

3 years ago

Verify 3/4 + (7/2 + -3/8) = 3/4 + 7/2 + (-3/8)

Mathematics

1 answer:

Iteru [2.4K]3 years ago

3 0

Answer:

when opening the bracket remember that the value outside the bracket will be used to multiply the value inside the bracket

You might be interested in

Evaluate the algebraic expression 6x-2 when<br> X=5/6

Neko [114]

Answer:

Step-by-step explanation:

Substitute x = $\frac{5}{6}$ into the expression

6x - 2

= 6 × $\frac{5}{6}$ - 2

= 5 - 2

= 3

4 0

3 years ago

please help with this it would mean a lot (you don't have to give the answer just help me understand it.) thanks

lord [1]

<span>2.065 IS THE ANSWER!!</span>

4 0

3 years ago

Is 143 and 63 relatively prime numberd

Igoryamba

Yes 143 and 63 are relatively prime because they have no prime factors in common

brainliest answer please?

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;

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

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

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