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Mazyrski [523]
2 years ago
8

HELP IM BEING TIOMEEEEEEEEDDDDDDDDDDDDDDDD

Mathematics
1 answer:
Anton [14]2 years ago
6 0

Answer:

-2.1

Step-by-step explanation:

x= 4/5

you put in 4/5 where x is supposed to be so you get: -2/5 - 4/5 + (-9/10)

-2/5 - 4-5 = -6/5

-6/5= -12/10

-12/10 + -9/10= -21/10 = -2.1

You might be interested in
F(x) = -9x + 2 and g(x) = -9x + 6, find (f - g)(7)
Lemur [1.5K]

Answer:

I think there is an error in the question because

(f-g) = -4

(f-g) (7) = NO SOLUTION

Step-by-step explanation:

f(x) = -9x + 2 \\g(x) = -9x + 6\\(f - g)(7)\\(f - g) = -9x + 2 - (-9x+6)\\(f - g) = -9x +2 +9x-6\\(f - g) = -9x +9x+2-6\\(f - g) = -4

7 0
2 years ago
Simplify and write answer in standard form.
mamaluj [8]

Answer:

x^3+8x^2+2x+1

Step-by-step explanation:

Combine like terms (terms with the same exponent).

x^3+(3x^2+5x^2)+(x+x)+1=x^3+8x^2+2x+1

"Standard form" means to list terms so the exponents go from high to low (descending order).

6 0
3 years ago
What is the economically efficient output level? 600 units 800 units 940 units 1160 units?
AURORKA [14]
Wot is the problu? its 940. no problem
8 0
2 years ago
Help please?!
Furkat [3]
Well from what i know here's what i would put........
-4 * -2 = 8 - 2 = 6
-4 * -1 = 4 - 2 = 2
-4 * 0 = 0 - 2 = -2
so it does represent a function......I really hope that helped!!

8 0
3 years ago
Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (
IrinaVladis [17]

Answer:

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14  and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median  of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = \dfrac{9+9}{2}

median = \dfrac{18}{2}

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = \dfrac{12+13}{2}

Median = \dfrac{25}{2}

Median = 12.5

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = Q3 - Q1

The interquartile range =  12.5 - 9

The interquartile range = 3.5

7 0
3 years ago
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