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3 years ago

Write the equation of the linear relationship in slope-intercept form, using decimals as needed.

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

7 0

Ok so look for the change between x and y.

patriot [66]3 years ago

3 0

The answer is y= 0.45x+3.5

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Find the constant variation for this equation. R=.51/d^2

nadya68 [22]

A constant of variation is part of a direct relation - linear function. The equation given is an inverse relation.
It does not have a constant of variation.

7 0

4 years ago

Chris puts one tenth of each paycheck into savings. Her current paycheck is $200. She decides to also put into savings, any amou

lions [1.4K]

Using given information we have to fill the blank spaces for respective parts:

Part(A): Original ratio of saved to total money per paycheck

Answer: $\frac{20}{200}$

Explanation: Original payment was 200 and saving amount is 1 tenth which is 200*(1/10)=20. Hence ratio is 20/200

Part(B): Final ratio of saved to total money per paycheck___________________

Answer: $\frac{x+20}{x+200}$

Explanation: He also saved money more than 200 represented by x so the new ratio becomes (x+20)/(x+200)

Part(C): Final amount saved per paycheck______________________________

Answer: x+20

Explanation: Final saving is just 20 and extra money over 200 so final saving is (x+20).

8 0

4 years ago

Express 0.055 in scientific notation.

aalyn [17]

5x10^-2 I believe is your answer

3 0

3 years ago

Read 2 more answers

The force pulling a truck downhill is 2,000N, the mass of the truck is 40,000 kg. What is the acceleration?

JulijaS [17]

For this case we have that Newton's second law is given by:

$F = m * a$

Where:

m: It's the mass

a: Acceleration

F: It's the force

According to the data we have:

$2,000 = 40,000 * a$

Dividing between 40,000 on both sides of the equation:

$a = \frac {2,000} {40,000}\\a = 0.05$

Thus, the acceleration is $0.05 \frac {m} {s ^ 2}$

Answer:

$0.05 \frac {m} {s ^ 2}$

7 0

4 years ago

Find the two x-intercepts of the function f and show that f '(x) = 0 at some point between the two x-intercepts.

suter [353]

Step-by-step explanation:

f(x) = x√(x+8)

When f(x) = 0,

0 = x√(x+8)

x = 0, x + 8 = 0

x = -8

f'(x) = (1)√(x+8) + x • ½(x+8)^(-½) = 0

0 = √(x+8) + 1/[2√(x+8)] (x)

-2(x+8) = x

-2x - 16 = x

x = -16/3

(-8 < -16/3 < 0)

Therefore, x-intercept of f'(x) = 0 is somewhere between the two x-intercepts, ranging from -8 to 0.

7 0

3 years ago