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2 years ago

Will GIVE BRAINILEST TO WHO ANSWERS Subtract 9x^2+7x-29x 2 +7x−2 from 8x-108x−10.

Mathematics

2 answers:

Rudik [331]2 years ago

7 0

Answer:

−9x

+x−8

Step-by-step explanation:

gogolik [260]2 years ago

5 0

Answer:

20x^2 - 107x - 10.

Step-by-step explanation:

8x - 108x - 10 - (9x^2 + 7x -29x^2)

simplifying:

-100x - 10 - (7x - 20x^2)

20x^2-100x - 7x - 10

20x^2 - 107x - 10

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8. The expression below is evaluated when x = 9, y = 3,

Reptile [31]

Answer:

Step-by-step explanation:

First of all substitute the values to where ever you see their corresponding variables to make them multiply.

x = 9

y = 3

z = 2

so we get

8x as 8(9),

7z as 7(2)

2y as 2(3)

then the it's will be

8(9) - 7(2) + 2(3)

which is

72 - 14 + 6

finally we get

4 0

1 year ago

The box plots show the math scores of students in two different classes. Based on the box plots, which statement is correct?

Allisa [31]

Based on the box plots, the statement which is correct is that: A. The median score of Class A is greater than the median score of Class B.

<h3>What is a box and whisker plot?</h3>

In Mathematics, a box plot is also referred to as box and whisker plot and it can be defined as a type of chart that can be used to graphically or visually represent the five-number summary of a data set with respect to locality, skewness, and spread.

Additionally, the five-number summary of any box plot (box and whisker plot) include the following:

Minimum
First quartile
Median
Third quartile
Maximum

By critically observing the box plot (box and whisker plot) which represent the math scores of students in in two different classes, we can reasonably and logically deduce the following median scores;

Median score of class A = 80

Median score of class B = 75

Therefore, a median score of 80 in Class A is greater than the median score of 75 in Class B.

Read more on box plots here: brainly.com/question/14277132

#SPJ1

6 0

4 months ago

Eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Gekata [30.6K]

Answer:

Option 1 and Option 4

Step-by-step explanation:

The rest are just wrong.

6 0

2 years ago

Read 2 more answers

What is the average of 1/2 and 1/3

nadya68 [22]

Hi there!

In order to fin the average, you need to add all the terms together and then divide the result by the number of terms :

( $\frac{1}{2} + \frac{1}{3}$ ) ÷ 2 = average

Now, I’m assuming that you know that in order to add fractions, both fractions must have the same denominator (bottom number in a fraction). Since these fractions do not have the same denominator, we must give them one.

To find the lowest common denominator (which will help us solve this problem), we must find out what 2 & 3 go into. Well, both numbers go into 6!

So, if the denominator is now 6, you must multiply the numerator (numbers about the “/” line in this case are both 1) by how much you multiplied its denominator by.

For 1/2, you multiplied 2 by 3 to get 6. Therefore, you must multiply the 1 by 3 aswell.

for 1/3, you multiplied 3 by 2 to get 6. Therefore, you must multiply the 1 by 2.

and now you have 3/6+2/6 since the denominators are the same, you can add the numerators normally which gives you 5/6

$\frac{5}{6}$ ÷ 2 = $\frac{5}{12}$

Your answer is : $\frac{5}{12}$

There you go! I really hope this helped, if there's anything just let me know! :)

3 0

2 years ago

a jar contains 5 blue marbles and 3 red marbles. suppose you choose a marble at random and do not replace it. then choose a seco

lina2011 [118]

$\mathbb P(R_1\cap R_2)=\mathbb P(R_2\mid R_1)\mathbb P(R_1)$

The probability of drawing a red marble on the first attempt is

$\mathbb P(R_1)=\dfrac{\dbinom31}{\dbinom51}=\dfrac35$

After the first marble is drawn, and we know it to be red, we're drawing the next marble from a pool with one less marble than before (i.e. conditioning on the event $R_1$ ). So

$\mathbb P(R_2\mid R_1)=\dfrac{\dbinom21}{\dbinom71}=\dfrac27$

And so

$\mathbb P(R_1\cap R_2)=\dfrac35\cdot\dfrac27=\dfrac6{35}$

4 0

3 years ago