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3 years ago

1. a single standard number cube is rolled, what is the probability of getting 4 or 5?

1 answer:

5 0

There are 6 sides on a standard number cube. Numbers are listed on it from 1-6. This being said, 4 and 5 together make up 2/6 of the cube. Since this fraction can be reduced, this leaves us with C. 1/3

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Help Please, anyone good at domain values of composite functions.

Delicious77 [7]

Answer:

second choice

Step-by-step explanation:

$(f \circ g)(x) =$

$= f(g(x))$

$= f(-7x + 7)$

$= \sqrt{-7x + 7 + 3}$

$= \sqrt{-7x + 10}$

Let x = 5:

$(f \circ g)(5) = \sqrt{-7(5) + 10}$

$(f \circ g)(5) = \sqrt{-35 + 10}$

$(f \circ g)(5) = \sqrt{-25}$

Since using x = 5 gives the square root of a negative number, 5 is not in the domain of the composition of the functions.

Answer: second choice

3 0

3 years ago

Prove: Quadrilateral ABCD is a parallelogram.

Brilliant_brown [7]

Answer:

Step 3.

m∠ AEB = m∠ CED .........By Vertical Angles Theorem.

Step-by-step explanation:

Vertical Angles Theorem:

Vertical angle theorem states that vertical angles, angles that are opposite each other and formed by two intersecting lines,are congruent.

If two lines intersect each other we have two pair of vertical opposite angles. As shown in the figure.

Here,

∠ 1 and ∠ 2 are vertical opposite angles and also they are equal.

∠ 3 and ∠ 4 are also vertical opposite angles and also they are equal.

For,

step 3. m∠ AEB = m∠ CED

Therefore, the reason for step 3 of this proof is Vertical Angles Theorem.

6 0

3 years ago

Solve each equation and leave the answer in terms of “i”

almond37 [142]

17)
x² + 8 = -8
x² = -8 - 8
x² = 16
x = ±√-16
x = ±√16i
x = <span>±4i
</span>
18)
x² + 5 = -3
x² = -3 - 5
x² = -8
x = ±√-8
x = ±√8i
x = ±2√2i

19)
x² + 3 = 0
x² = -3
x = ±√-3
x = <span>±</span>√3i

hope this helps, God bless!

3 0

3 years ago

Which of these best describes a scientific theory

Sidana [21]

Answer:

Step-by-step explanation:

Have a good day i hope this helps :)

4 0

3 years ago

Rationalise the denominator of:<br>1/(√3 + √5 - √2)

Paul [167]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \: = \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \: = \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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<h3><u>More Identities to </u><u>know:</u></h3>

$\purple{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\purple{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

6 0

3 years ago