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3 years ago

Can someone please answer 1 for me? Whoever can answer gets brainliest

Mathematics

1 answer:

GenaCL600 [577]3 years ago

6 0

ANSWER
Day 1 he gets $7
Day 2 he gets $14
Day 3 he gets $21

EXPLANATION:

add 7 to each day!
Hope that helps <3

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I need a answer asap Please!!

Ganezh [65]

Answer:

The ordered pairs (3 , 6) , (5 , 10) show a proportional relationship ⇒ last answer

Step-by-step explanation:

* Lets explain how to sole the problem

- Proportional relationship describes a simple relation between

two variables

- In direct proportion if one variable increases, then the other variable

increases and if one variable decreases, then the other variable

decreases

- In inverse proportion if one variable increases, then the other variable

decreases and if one variable decreases, then the other variable

increases

- The ratio between the two variables is always constant

- Ex: If x and y are in direct proportion, then x = ky, where k

is constant

If x and y in inverse proportion, then x = k/y, where k is constant

* Lets solve the problem

# Last table

∵ x = 3 and y = 6

∴ x/y = 3/6 = 1/2

∵ x = 5 and y = 10

∴ x/y = 5/10 = 1/2

∵ 1/2 is constant

∵ x/y = constant

∴ x and y are proportion

* The ordered pairs (3 , 6) , (5 , 10) show a proportional relationship

7 0

3 years ago

Find the solution of the separable equations a) the general solution: displaystyle dy/dt = t/[y^2] b) the unique solution to: dy

Xelga [282]

Answer:

(a) The general solution of given differential equation is $\frac{y^3}{3}=\frac{t^2}{2}+C$ .

(b) The unique solution is $\frac{y^2}{2}=t-\frac{5}{2}$ .

Step-by-step explanation:

(a)

The given differential equation is

$\frac{dy}{dt}=\frac{t}{y^2}$

Use variable separable method, to solve the above differential equation.

Separate the variables.

$y^2dy=tdt$

Integrate both sides.

$\int y^2dy=\int tdt$

$\frac{y^3}{3}=\frac{t^2}{2}+C$

The general solution of given differential equation is $\frac{y^3}{3}=\frac{t^2}{2}+C$ .

(b)

The given differential equation is

$\frac{dy}{dt}=\frac{1}{y}$

Use variable separable method, to solve the above differential equation.

Separate the variables.

$ydy=1dt$

Integrate both sides.

$\int ydy=\int 1dt$

$\frac{y^2}{2}=t+C$ ... (1)

It is given that y=1 at t=3. Substitute y=1 and t=3 in the above equation.

$\frac{(1)^2}{2}=(3)+C$

$\frac{1}{2}-3=C$

$-\frac{5}{2}=C$

Substitute $C=-\frac{5}{2}$ in equation (1).

$\frac{y^2}{2}=t-\frac{5}{2}$

Therefore the unique solution is $\frac{y^2}{2}=t-\frac{5}{2}$ .

6 0

4 years ago

Which quadratic rule represents the data in the table?

Vaselesa [24]

Answer:

x^2 - 5

Step-by-step explanation:

x = -1, y = 1 - 5 = -4.

x = 0, y = 0-5 = -5.

x = 1, y = 1 - 5 = -4.

7 0

3 years ago

Can you help me find the distance between each point round your answer to the nearest tenth

blagie [28]

In the first graph it’s graphed to be nearest 10

5 0

3 years ago

Determine the measure of each angle using the picture below. Assume the two lines are parallel.

marta [7]

Answer:

Step-by-step explanation:

Angle 5 is opposite to 37 => It's equal to 37.
Angle 6 and 7 are supplementary to angle 5 => They are equal to 180 - 37 = 143 each.

Angle 4 is corresponding to Angle 7, and since the 2 lines are parallel, it's equal to 143.

Angle 1 and 3 is supplementary to Angle 4, so they are equal to 180-143 = 37 each

Angle 2 is opposite to Angle 4, so it's equal to 143.

Answer : Angle 1,3,5 = 37
Angle 2,4,6,7 = 143

8 0

3 years ago