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3 years ago

12

Using graph paper, determine the line described by the given point and slope. Click to show the correct graph below. (0, 6) and

-1

1 answer:

guapka [62]3 years ago

3 0

(0, 6) and slope = -1

y = -x + b

b = y + x

b = 6 + 0

b = 6

so equation

y = -x + 6

when x = 1 ,y = -1 + 6 = 5

when x = 0 ,y = 6

Answer is the second graph

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999+6666+8888+66654+66=

scoundrel [369]

Answer:

83273

Step-by-step explanation:

8 0

3 years ago

Nine squared plus a number (<br> b. is 89.

Andrei [34K]

The number is 8 since 9 squared is 81

4 0

3 years ago

A system of linear equations includes the line that is created by the equation y = x+ 3, graphed below, and the line through the

ad-work [718]

Answer:

(8,11)

Step-by-step explanation:

1st line

y = x+ 3

We need to find the equation of the other line

We have two points (3, 1) and (4, 3)

Slope

m = (y2-y1)/(x2-x1)

= ( 3-1)/(4-3)

= 2/1 = 2

The slope intercept form is y = mx+b where m is the slope and b is the y intercept

y =2x+b

Using the point (4,3)

3 = 2(4)+b

3 = 8+b

3-8=b

-5 =b

y = 2x-5

We have 2 lines

y =x+3 and y = 2x-5

Setting them equal to each other

x+3 = 2x-5

Subtract x from each side

x+3-x = 2x-5-x

3 = x-5

Add 5 to each side

3+5 =x-5+5

8=x

Now we can find y

y =x+3

y = 8+3

y=11

7 0

3 years ago

The crayon balloon is made up of a cone and a cylinder. What is the volume, in cubic inches, of the crayon balloon?

gavmur [86]

Answer:

A. 69.12 in.³

Step-by-step explanation:

Volume of the crayon balloon = volume of the cone + volume of the cylinder

= ⅓πr²h + πr²h

✔️Volume of Cone = ⅓πr²h

r = ½(2) = 1 in.

h = 3 in.

Volume of cone = ⅓*π*1²*3 = 3.142 in.³

✔️Volume of Cylinder = πr²h

r = ½(2) = 1 in.

h = 21 in.

Volume of Cylinder = π*1²*21 = 65.97 in.³

✅Volume of the crayon balloon = 3.14 + 65.973 = 69.115

= 69.12 in.³ (nearest hundredth)

6 0

3 years ago

Find the solution of the initial value problem<br><br> 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