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2 years ago

Find the slope of the line that passes through the given points (9, -3) and (9, 4)

Mathematics

1 answer:

EleoNora [17]2 years ago

6 0

Answer: No slope.

Step-by-step explanation:

No slope, the x values are the same yet they have different Y values so there isn't a slope and this isn't a function.

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The measure of five of of the interior angles of a hexagon are 150, 100, 80, 165, and 150. what is the measure of the sixth inte

givi [52]

Answer:

75°

Step-by-step explanation:

The sum of the interior angles of a polygon is

sum = 180° (n - 2) ← n is the number of sides

Here n = 6 , then

sum = 180° × 4 = 720°

let x be the sixth angle, then sum and equate to 720°

150 + 100 + 80 + 165 + 150 + x = 720

645 + x = 720 ( subtract 645 from both sides )

x = 75

The sixth interior angle is 75°

7 0

2 years ago

What is the sum of the areas of circle C and circle D?

Yanka [14]

Answer:

98pi

Step-by-step explanation:

Area of circle: pi(r)^2

A1=pi(7)^2

A2=pi(7)^2

A2+A1=98pi

4 0

2 years ago

Solve for h in A = ½ ah - ½ bh. Solve for h in A = ½ ah - ½ bh .

pantera1 [17]

Answer:

h = A / ( 1/2 a - 1/2 b)

Step-by-step explanation:

A = ½ ah - ½ bh

Factor out an h

A = h( 1/2 a - 1/2 b)

Divide each side by ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h ( 1/2 a - 1/2 b)/ ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h

h = A / ( 1/2 a - 1/2 b)

7 0

1 year ago

Quick Interest Rate Math help please !! :0

lara [203]

Answer : r = 4$

hope this helps

8 0

2 years ago

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

7 0

2 years ago