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2 years ago

Apply the distributive property to factor out the greatest common factor. 4+10=

Mathematics

1 answer:

almond37 [142]2 years ago

4 0

4 + 10 = 2 x 2 + 2 x 5 = 2 x (2 + 5)

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10-d = -34-5d solve this equation

algol13

<h3>Explanation</h3>

Given the equation.

$10 - d = - 34 - 5d$

Solve the equation for d-term.

Because we cannot subtract or add up constants and variables, we simply move the same variable term to the same side and constant term to the same constant side.

$10 + 34 = - 5d + d \\ 44 = - 4d \\ \frac{44}{ - 4} = d \\ - 11 = d$

Answer Check

Substitute d = -11 in the equation.

$10 - ( - 11) = - 34 - 5( - 11) \\ 10 + 11 = - 34 + 55 \\ 21 = 21$

The equation is true for d = -11.

<h3>Answer</h3>

 $\large \boxed{d = - 11}$ 

7 0

3 years ago

Read 2 more answers

What is the basic ratio for 30:42

lianna [129]

30:42 is represented as 30/42
The lowest form for that is 5/7. Hope that helped. Good luck

6 0

3 years ago

Which ordered pair will solve the equation 2 × 4 + x = y?

Bezzdna [24]

This answer uses NMF, which you can find out about on my profile:

Preliminary work:
Following the BIDMAS order of operations, we can calculate part of it already, and that's the 2•4, which equals 8.
Therefore, the equation now reads:
8+x = y

x = 5:
8+5 = 13
13 ≠ 16
13 ≠ y

x = 4:
8+4 = 12
12 = 12
12 = y

Therefore, the pair is (4, 12)

5 0

3 years ago

Great tennis players use Hexrackets. There if you use a Hexracket, you are a great tennis nyer.

34kurt

Answer:

The conclusion is invalid.

The required diagram is shown below:

Step-by-step explanation:

Consider the provided statement.

Great tennis players use Hexrackets. Therefore, if you use a Hexracket, you are a great tennis player.

From the above statement we can concluded that Great tennis players use Hexrackets. But it may be possible that some people who use a haxracket are not great tennis player.

Therefore, the conclusion is invalid.

The required diagram is shown below:

8 0

3 years 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:

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