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2 years ago

Find the value of g(x)=2-x^2 and f(x)=2x-1 , a, g(-8)

Mathematics

2 answers:

Setler [38]2 years ago

4 0

Answer:

-62

Step-by-step explanation:

g(x)=2-x^2

Let x = -8

g(-8)=2-(-8)^2

= 2- 64

= -62

masya89 [10]2 years ago

3 0

Solve for the function of g then use what you get for g to solve for f.

g(-8) = 2 - (-8)^2

g(-8) = 2 - 64

g(-8) = -2

f(-2) = 2(-2) - 1

f(-2) = -4 - 1

f(-2) = -5

Best of Luck!

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Resistors for electronic circuits are manufactured on a high-speed automated machine. The machine is set up to produce a large r

Anettt [7]

Answer:

See explanation

Step-by-step explanation:

Given

See attachment for proper presentation of question

Required

Mean and Range

To do this, we simply calculate the mean and the range of each row.

$\bar x = \frac{\sum x}{n}$ ---- mean

Where:

$n = 4$ ---- number of rows

$R = Highest - Lowest$ --- range

So, we have:

Sample 1

$\bar x_1 = \frac{1027+ 994 +977 +994 }{4}$

$\bar x_1 = 998$

$R_1 = 1027- 994$

$R_1 = 33$

Sample 2

$\bar x_2 = \frac{975 +1013 +999 +1017}{4}$

$\bar x_2 = 1001$

$R_2 = 1017 - 975$

$R_2 = 42$

Sample 3

$\bar x_3 = \frac{988 +1016 +974 +997}{4}$

$\bar x_3 = 993.75$

$R_3 = 1016-974$

$R_3 = 42$

Sample 4

$\bar x_4 = \frac{998 +1024 +1006 +1010}{4}$

$\bar x_4 = 1009.5$

$R_4 = 1024 -998$

$R_4 = 26$

Sample 5

$\bar x_5 = \frac{990 +1012 +990 +1000}{4}$

$\bar x_5 = 998$

$R_5 = 1012 -990$

$R_5 = 22$

Sample 6

$\bar x_6= \frac{1016 + 998 +1001 +1030}{4}$

$\bar x_6= 1011.25$

$R_6= 1030-998$

$R_6= 32$

Sample 7

$\bar x_7 = \frac{1000 +983 +979 +971}{4}$

$\bar x_7 = 983.25$

$R_7 = 1000-971$

$R_7 = 29$

Sample 8

$\bar x_8 = \frac{973 +982 +975 +1030}{4}$

$\bar x_8 = 990$

$R_8 = 1030-973$

$R_8 = 57$

Sample 9

$\bar x_9 = \frac{992 +1028 +991 +998}{4}$

$\bar x_9 = 1002.25$

$R_9 = 1028 -991$

$R_9 = 37$

Sample 10

$\bar x_{10} = \frac{997 +1026 +972 +1021}{4}$

$\bar x_{10} = 1004$

$R_{10} = 1026 -972$

$R_{10} = 54$

Sample 11

$\bar x_{11} = \frac{990 +1021 +1028 +992}{4}$

$\bar x_{11} = 1007.75$

$R_{11} = 1028 -990$

$R_{11} = 38$

Sample 12

$\bar x_{12} = \frac{1021 +998 +996 +970}{4}$

$\bar x_{12} = 996.25$

$R_{12} = 1021 -970$

$R_{12} = 51$

Sample 13

$\bar x_{13} = \frac{1027 +993 +996 +996}{4}$

$\bar x_{13} = 1003$

$R_{13} =1027 -993$

$R_{13} =34$

Sample 14

$\bar x_{14} = \frac{1022 +981 +1014 +983}{4}$

$\bar x_{14} = 1000$

$R_{14} = 1022 -981$

$R_{14} = 41$

Sample 15

$\bar x_{15} = \frac{977 +993 +986 +983}{4}$

$\bar x_{15} = 984.75$

$R_{15} = 993-977$

$R_{15} = 16$

8 0

2 years ago

P + 13x = q p = 7 x = 4 q =

Leviafan [203]

13(4)=52 52+7=59 so q=59

5 0

3 years ago

A School decides to organize field trips for all students. Tickets for first years were sold at GH¢0.40 per student and continui

Pepsi [2]

Answer:

60%

Step-by-step explanation:

You can solve this problem by setting up a system of equations.

Let's say that the number of tickets bought by students in the first year is x, and the number bought by continuing students is y. From there, you can set it up like this:

0.4x+0.2y=160

x+y=500

Now, you can multiply the first equation by 5 on both sides to get:

2x+y=800

Subtracting the second equation from the first equation now yields:

x=300

y=200

Since 300 of the 500 tickets bought were from the first year students, and 300/500 is 0.6, 60% of the students who bought the ticket were first year students. Hope this helps!

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3 years ago

You rent an apartment that costs $900 per month during the first year, but the rent is

olchik [2.2K]

Answer:1493.1

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

The depth of a lake in Lexington changes over time due to rainfall and evaporation. A few months ago, the depth was 1,300 feet.

SVETLANKA909090 [29]

Answer:

The Depth of the lake had increased by 19%.

Step-by-step explanation:

Given:

Depth of lake few months ago = 1300 ft

depth of lake currently = 1547 ft

We need to find the percent of increase in depth of lake.

Solution:

First we will find the increase in depth of lake.

Increase in depth of lake can be calculated by subtracting Depth of lake few months ago from depth of lake currently.

framing in equation form we get;

increase in depth of lake = $1547-1300= 247\ ft$

Now to find the percent of increase in depth of lake we will divide increase in depth of lake from Depth of lake few months ago and then multiply by 100.

framing in equation form we get;

percent of increase in depth of lake = $\frac{247}{1300}\times 100 = 19\%$

Hence the Depth of the lake had increased by 19%.

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3 years ago