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Arisa [49]
3 years ago
7

Find an expression which represents the difference when (10x+6y) is subtracted from (6x-8y) in simplest terms

Mathematics
2 answers:
Anna007 [38]3 years ago
6 0

Answer:

x−14y

Step-by-step explanation:

(7x−8y)+(−6x−6y)

The "sum" is the result of addition

7x-8y-6x-6y

7x−8y−6x−6y

Remove parentheses

7x−8y−6x−6y

Identify like terms

7x−6x−8y−6y

Commutative property

7x−6x  

−14y

−8y−6y

​

Combine like terms

1x−14y

x−14y

Coefficient of 1 isn't needed

Tema [17]3 years ago
3 0

Answer:

4x+2y

Step-by-step explanation:

You might be interested in
Cuál será la razón entre la cantidad de lejía y la cantidad de agua, si la taza contiene 200 ml (mililitros) y cada litro es 100
tatyana61 [14]

Answer:

Ratio of bleach and water = 1:4

Relación de lejía y agua = 1:4

Step-by-step explanation:

English Translation

What will be the ratio between the amount of bleach and the amount of water, if the filled, 1 L cup contains 200 ml (milliliters) of bleach and each liter is 1000 ml (milliliters)

The cup contains 1000 mL of liquid.

The cup contains 200 mL of bleach and given that the rest is water, there are (1000 - 200) mL of water in the cup.

Amount of bleach in cup = 200 mL

Amount of water in cup = 800 mL

Ratio of bleach to water = 200:800

divide by the least number

Ratio of bleach to water = 1:4

In Spanish/En español

La taza contiene 1000 ml de líquido. La taza contiene 200 ml de lejía y dado que el resto es agua, hay (1000 - 200) ml de agua en la taza.

Cantidad de lejía en taza = 200 ml

Cantidad de agua en taza = 800 ml

Relación de lejía a agua = 200:800

dividir por el menor número

Relación de lejía a agua = 1:4

Hope this Helps!!!

8 0
3 years ago
A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 200 times. Here are the observed f
Alona [7]

Solution :

The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.

Null hypothesis, $H_0:p_1=p_1=p_3=p_4=p_5=p_6=\frac{1}{6}$

That is the loaded die behaves as a fair die.

Alternative hypothesis, $H_a$ : loaded die behave differently than the fair die.

Number of attempts , n = 200

Expected frequency, $E_i=np_i$

                                        $=200 \times \frac{1}{6} = 33.333$

Test statistics, $x^2= \sum^6_{i=1} \frac{(O_i-E_i)^2}{E_i} $

                            $=\frac{(28-33.333)^2}{33.333}+\frac{(29-33.333)^2}{33.333}+\frac{(40-33.333)^2}{33.333}+\frac{(41-33.333)^2}{33.333}+\frac{(28-33.333)^2}{33.333}+$$\frac{(34-33.333)^2}{33.333}$

≈ 5.8

Degrees of freedom, df = n - 1

                                       = 6 - 1

                                       = 5

Level of significance, α = 0.10

At α = 0.10 with df = 5, the critical value from the chi square table

$x^2_{\alpha}= \text{chi inv}(0.10,5)$

     = 9.236

Thus the critical value is $x_{\alpha}^2=9.236$

$P \text{ value} = P[x^2_{df} \geq x^2]$

             $=P[x^2_5\geq 5.80]$

             = chi dist (5.80, 5)

             = 0.3262

Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject $H_o$ at 10% LOS.

Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.

3 0
3 years ago
Samir bought 3 pounds of cement to repair the crack in his sidewalk each crack needs to be filled with 2 oz of cement part a how
krek1111 [17]
Uhm let’s see how about part b
7 0
3 years ago
Justin has 20 pencils 25 erasers and 40 paperclip he organize them into groups with the same number of items in each group all t
e-lub [12.9K]
Find the GCF (Greatest common factor) of all the numbers.
There will be 4 pencils. 5 erasers. 8 paper clips. Hope I could help!
Btw- the GCF is 5.
8 0
3 years ago
Collin is building a deck on the back of his family‘s house he has enough lumber for the deck to be 144 ft.² the length should b
stira [4]
18x8
Use a calculator and plug in numbers
6 0
3 years ago
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