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7nadin3 [17]
3 years ago
12

The functions q and r are defined as follows.

Mathematics
1 answer:
UkoKoshka [18]3 years ago
3 0
Q(3) = -2(3)+2 = -4
r(q(3)) = (-4)^2-1 = 15
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5 4/8 + 1 3/8 =<br> 1. 33/8<br> 2. 55/8<br> 3. 6 7/16
Dovator [93]

Answer: 2. 55/8

Step-by-step explanation: Add the whole numbers first (5+1 = 6), then add the fractions (4/8 + 3/8 = 7/8). Now we have 6 7/8 and since they want the answer in fraction form we will multiply the 6 and the 8 (6x8=48) and then add the 7 to the 48 (7+48 =55) and we leave our answer over the denominator which is 8. This leaves us with the final answer of 55/8

4 0
1 year ago
(0,0), (1,5), (2,20), (3,45), (4,80) what's the rule?
yanalaym [24]
Hmm, seems to multiply by 5
wait, no

first is multiply by 5
2nd is multiply by 10
3rd is multiply by 15
4th is multiply by 20
seems to be

f(x)=x(5x) or f(x)=5x^2
the rule is square the input then multiply by 5
5 0
3 years ago
6) The scatter plot below shows the temperatures (y), in degrees Fahrenheit (°F), that were recorded at different altitudes (x),
prohojiy [21]

Answer:

y = − (7 /2) x + 60 equation could represent the line of best fit for the temperatures, in degrees Fahrenheit, based on the altitudes, in thousands of feet

Step-by-step explanation:

Given:

A plot for temperature and altitude

(Refer the attachment)  .......FOR GRAPH

To Find:

Correct relationship between them

Solution:

By using  given relationship as we can conclude.

1)y=-5x+70:

Put y=0 then

5x=70

x=70/5

x=14

This value dont corresponds when y=0 in graph

2) y=-10x+70

put y=0 then

10x=70

x=7

This value dont corresponds when y=0 in graph very less than original value.

3)y=-(7/2)x+60

put y=0 then

(7/2)x=60

7x=120

x=120/7

x=17.14

This value  corresponds when y=0 in graph

4)y=-(9/2)x+60

Put y=0 then

(9/2)x=60

9x=120

x=120/9

x=13.33

This value dont corresponds when y=0 in graph which much less than original value.

7 0
3 years ago
X^2-6x=7 completing the square
kozerog [31]

Answer:

x = 7, -1

Step-by-step explanation:

Use the formula (b/2)^2 in order to create a new term. Solve for x by using this term to complete the square.

3 0
3 years ago
Iridium -192 has a half life of 74 days after 148 days, how many milligrams of 900 mg sample will remain?
Ad libitum [116K]
\bf \textit{Amount for Exponential Decay using Half-Life}&#10;\\\\&#10;A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{initial amount}\to &900\\&#10;t=\textit{elapsed time}\to &148\\&#10;h=\textit{half-life}\to &74&#10;\end{cases}&#10;\\\\\\&#10;A=900\left( \frac{1}{2} \right)^{\frac{148}{74}}\implies A=900\left( \frac{1}{2} \right)^2\implies A=225
8 0
3 years ago
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