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2 years ago

Write in slope intercept form

Mathematics

2 answers:

ololo11 [35]2 years ago

5 0

option b because the y-intercept is at -5, so A and B are left. then since the line of the graph is going down, from left to right, your slope is going to be a negative. B is your final answer :)

saveliy_v [14]2 years ago

3 0

Answer:

option B

Step-by-step explanation:

the slope is -2, so the equation must have -2x in it.

This narrows it down to option C and B

the intercept is at -5, not +5 so the correct answer is y = -2x - 5

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Ten less than half of a number(x) is 6.

Likurg_2 [28]

Answer:

x = 32

Step-by-step explanation:

Given the following data;

Unknown number = x

Translating the word problem into an algebraic equation, we have;

$\frac {x}{2} - 10 = 6$

Lowest common denominator (LCD) = 2

We multiply all through by 2;

$2* \frac {x}{2} - 2*10 = 2*6$

$x - 20 = 12$

$x = 12 + 20$

x = 32

Therefore, the unknown number is 32.

3 0

2 years ago

HELP ASAP PLEASE!!!!!!

Nuetrik [128]

Answer:

Step-by-step explanation:

The common ratio here is 3. Note that 4 is the first term of this sequence, and that n is the index: first term, second term, etc.

Every term is found by mult. the previous term by 3.

3 0

3 years ago

Consider the quadratic function f(y) = 8y2 – 7y + 6. What is the constant of the function?

Paraphin [41]

The constant of the function would be 6.
Hope this helps!

5 0

3 years ago

Read 2 more answers

Solve the initial-value problem using the method of undetermined coefficients.

andrey2020 [161]

First check the characteristic solution. The characteristic equation to this DE is

r ² - r = r (r - 1) = 0

with roots r = 0 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(0x) + C₂ exp(1x)

y (char.) = C₁ + C₂ exp(x)

For the particular solution, we try the ansatz

y (part.) = (ax + b) exp(x)

but exp(x) is already accounted for in the second term of y (char.), so we multiply each term here by x :

y (part.) = (ax ² + bx) exp(x)

Differentiate this twice and substitute the derivatives into the DE.

y' (part.) = (2ax + b) exp(x) + (ax ² + bx) exp(x)

… = (ax ² + (2a + b)x + b) exp(x)

y'' (part.) = (2ax + 2a + b) exp(x) + (ax ² + (2a + b)x + b) exp(x)

… = (ax ² + (4a + b)x + 2a + 2b) exp(x)

(ax ² + (4a + b)x + 2a + 2b) exp(x) - (ax ² + (2a + b)x + b) exp(x)

= x exp(x)

The factor of exp(x) on both sides is never zero, so we can cancel them:

(ax ² + (4a + b)x + 2a + 2b) - (ax ² + (2a + b)x + b) = x

Collect all the terms on the left side to reduce it to

2ax + 2a + b = x

Matching coefficients gives the system

2a = 1

2a + b = 0

and solving this yields

a = 1/2, b = -1

Then the general solution to this DE is

y(x) = C₁ + C₂ exp(x) + (1/2 x ² - x) exp(x)

For the given initial conditions, we have

y (0) = C₁ + C₂ = 6

y' (0) = C₂ - 1 = 5

and solving for the constants here gives

C₁ = 0, C₂ = 6

so that the particular solution to the IVP is

y(x) = 6 exp(x) + (1/2 x ² - x) exp(x)

3 0

2 years ago

I don't understand what it's ask or how to do it .

k0ka [10]

6:1 , basically asking the ratio of one to the other. Understand?

7 0

2 years ago

Read 2 more answers