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2 years ago

How did i respond the question number 108

Mathematics

1 answer:

mafiozo [28]2 years ago

4 0

You can multiply each fraction on both top and bottom so that they have 24 as the denominator. The equation becomes $\frac{20}{24} m- \frac{10}{24} = \frac{21}{24} m+ \frac{16}{24}$
subtract 16/24 to both sides and you get
$\frac{20}{24} m- \frac{26}{24} = \frac{21}{24} m$
now subtract 20/24 from both sides and
$\frac{-26}{24} = \frac{21}{24} m$ 
multiply 24 on both sides
-26=21m
and divide by 21 on both sides
m=-26/21 or -1.24

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Solve 2(x + 1) = 2x + 5.

BlackZzzverrR [31]

Answer:

No solutions

Step-by-step explanation:

2(x+1)=2x+5

Expand the parentheses:

2x+2=2x+5

Subtract 2x from both sides:

2=5

Since this is not true, this equation has no solutions. Hope this helps!

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3 years ago

Read 2 more answers

One baseball team played 36 games throughout their entire season. If this baseball team won 18 of those games, then what percent

gizmo_the_mogwai [7]

They have won 50% of their games.

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2 years ago

For this item, a non-integer answer should be entered as a fraction using / as the fraction bar.

Kipish [7]

The numerical expression, 2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3 = 67/24 on simplification using the BODMAS rule.

In the question, we are asked to simplify the numerical expression:

2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3.

To simplify the expression, we will follow the BODMAS rule, where B means Brackets, O means Of, D means Divide, M means Multiplication, A means Addition, and S means Subtraction.

2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3

= 2/3 ÷ 16 + (3/4 + 1/6) ÷ 1/3 {Solving 2⁴ = 16, before proceeding BODMAS}.

= 2/3 ÷ 16 + ((9+2)/12) ÷ 1/3 {Solving Brackets by taking LCM}

= 2/3 ÷ 16 + 11/12 ÷ 1/3 {Simplifying}

= 2/3 * 1/16 + 11/12 * 3/1 {Solving divisions by taking reciprocals}

= 1/24 + 11/4 {Multiplying}

= (1 + 66)/24 {Adding using LCM}

= 67/24 {Simplifying}.

Thus, the numerical expression, 2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3 = 67/24 on simplification using the BODMAS rule.

Learn more about the simplification of numerical expression at

brainly.com/question/17205434

#SPJ1

The provided question is incomplete. The complete question is:

"Type the correct answer in the box. Use numerals instead of words. For this item, a non-integer answer should be entered as a fraction using / as the fraction bar.

Simplify the numerical expression.

2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3

The expression has a value equal to."

5 0

1 year ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$