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3 years ago

Suppose that the functions r and s are defined for all real numbers x as follows. HELP PLEASE

Mathematics

1 answer:

Bas_tet [7]3 years ago

3 0

Answer:

-x-3, 7x-3, -15

Step-by-step explanation:

For first 2 just add/subtract their equations. For the last one (r.s)(-1) is the same as r(s(-1)) so s(-1)=4(-1)=-4 and r(-4)=3(-4)-3=-12-3=-15

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3 years ago

Jennifer works 3 1/2 hours each morning at the clinic. How many routine physicals could she complete in one morning? (Routine ph

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Answer:

Step-by-step explanation:

Total Work hours = 3 1/2 hours

One routine physical routine requires: 1/3 hours.

Number of routines can be completed = Total work hours / Time required for one physical routine

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In this case, we are unable to complete half a routine (0.5), so we will choose the highest possible whole number, in this case, its 10 routines.

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2 years ago

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3 years ago

Aisha wants to paint the four walls of her living room.

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The total amount Aisha would pay to paint her room is $54.

<h3>What is the cost of painting the rooms?</h3>

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Area of the three walls without a door : 3 x (length x width)

3 x (2.4 x 5.2) = 37.44 m²

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(2.4 x 5.2) - (2 x 0.8) = 10.88m²

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To learn more about how to calculate the area of a rectangle, please check: brainly.com/question/16595449

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2 years ago

Find the first term and the common ratio <br>third term =6 and seventh term= 96

mamaluj [8]

Answer:

First term a₁ = 3/2 and common ratio r = 2

Step-by-step explanation:

We need to find the first term and common ratio while we are given third term = 6 and seventh term = 96

Since common ratio is required so, the sequence is geometric sequence

The formula used is: $a_n= a_1r^{n-1}$

We are given: third term = 6 i,e

$a_3=a_1r(3-1)\\6=a_1r^2----eq(1)$

Seventh term = 96

$a_7=a_1r(7-1)\\96=a_1r^6----eq(2)$

Dividing eq(2) and eq(3)

$\frac{96}{6}=\frac{a_1r^6}{a1r2}\\16=\frac{r^6}{r^2}\\16=r^{6-2}\\=> r^4=16\\Taking \ fourth \ root\\ r=2,-2,2i,-2i\\ \ We \ will \ consider \ only \ positive \ value \ of \ r \ i.e \ \textbf{ r= 2}$

So, Common Ratio r = 2

Finding First term using eq(1)

$a_3=a_1r^2\\6=a_1(2)^2\\6=a_1(4)\\a_1= \frac{6}{4}\\a_1= \frac{3}{2}$

So, First term a₁ = 3/2 and common ratio r = 2

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3 years ago