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AysviL [449]
4 years ago
10

Raphael graphed the system of equations shown. y = – 3 y = x – 0.8 What is the best approximation for the solution to this syste

m of equations? (–3.2, –3) (–2.9, –3) (–2.2, –3) (–1.9, –3) Mark this and return
Mathematics
2 answers:
ira [324]4 years ago
6 0
The answer is C (-2.2, -3). 

In order to find the ordered pair we must use what we know from the first equation in the second. We already know that y = -3, so that goes into the second equation. 

y = x - 0.8
-3 = x - 0.8 
-2.2 = x
jonny [76]4 years ago
6 0
Given ---›

y = -3

&

y = x - 0.8

So,
=> x = y + 0.8

=> x = -3 + 0.8

=> x = -2.2

Therefore, the best approximation for the solution to this system of equations is = (–2.2, –3)
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The variable z is inversely proportional to x. When x is 6, z has the value 0.5. What is the value of z when x= 14?
Lilit [14]
Z is inversely proportional to X is the same as
z is directly proportional to 1/X
which can be written like this:-
y=k/X
________
0.5=k/6
0.5*6=k/6*6
k=3 ;
________
z = 3/14
Z = 0.2
7 0
3 years ago
Sarah is a computer engineer and manager and works for a software company. She receives a
daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = a

Given that:

a_4=129\\a_{10}=207

Formula for n^{th} term of an Arithmetic Progression is given as:

a_n=a+(n-1)d

Where d will represent the number of projects increased every year.

and n is the year number.

a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)

Subtracting (2) from (1):

78 = 6d\\\Rightarrow d =13

By equation (1):

129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90

<em>Number of projects in the first year = 90</em>

<em></em>

<em>b) </em>

Number of projects in the twelfth year =

a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233

Each project pays $500

Earnings in the twelfth year = 233 \times 500 = $116500

Sum of an AP is given as:

S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 \times 1938 = $969000

8 0
3 years ago
What's the full answer
Jet001 [13]
????????????????????????
8 0
3 years ago
You have been asked to cut a 1.5m roll of bubble wrap into 30 cm lengths, which are used to wrap CDs before posting to clients.
dlinn [17]

Answer:

c. 5

Step-by-step explanation:

1.5m = 1.5*100 cm = 150 cm

Divide 150 cm by 30 cm,

150/30 = 5

8 0
2 years ago
Write in standard notation 6.12 x 10 to the 3rd power
faltersainse [42]
6.12 x 10^3 = 6120 .........<span> in standard notation
</span><span>
hope it helps</span>
7 0
3 years ago
Read 2 more answers
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