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2 years ago

Write an equation of the line with the given y-intercept and the slope in slope-intercept form m=1/3; b=-1

Mathematics

1 answer:

Over [174]2 years ago

8 0

Answer:

y= 1/3x - 1

Step-by-step explanation:

1/3 is slope so m and -1 is the y-intercept

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Ill give a brainliest i need help its due today

aliya0001 [1]

Answer:

w=15

Step-by-step explanation:

-24=-3w+21

We move all terms to the left:

-24-(-3w+21)=0

We get rid of parentheses

3w-21-24=0

We add all the numbers together, and all the variables

3w-45=0

We move all terms containing w to the left, all other terms to the right

3w=45

w=45/3

w=15

8 0

3 years ago

Read 2 more answers

The minitab Stem and Leaf output that corresponds to a set of measurements is as follows: Stem?and?leaf of Survival N = 64 Leaf

dolphi86 [110]

Answer:

Step-by-step explanation:

Left skewed

2 and 3

8 0

2 years ago

Evaluate 7 + (-3x^2) For x = 0 A. -2 B. 0 C. 4 D. 7

Goryan [66]

Answer: D. 7

Step-by-step explanation:

The problem tells you that x = 0 so replace everything that is x in the equation with 0 so the problem becomes

7 + (-3(0)^2)

Now we just solve the problem normally.

solve the problem in the parenthesis first, we must multiply -3 times 0^2 is 0

so now our problem is 7 + 0 which is 7

8 0

3 years ago

Look at this shape:

worty [1.4K]

Answer:

In my opinion is B

4 0

2 years ago

Find the solution of the initial value problem dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago