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3 years ago

What is the solution to the inequality -6( p - 8) < -12?

Mathematics

2 answers:

Setler [38]3 years ago

6 0

Answer:

p>10

Step-by-step explanation:

\left(-6\left(p-8\right)\right)\left(-1\right)>\left(-12\right)\left(-1\right)

6\left(p-8\right)>12

\frac{6\left(p-8\right)}{6}>\frac{12}{6}

p-8>2

p-8+8>2+8

p>10

coldgirl [10]3 years ago

3 0

Answer:

p > 10

Step-by-step explanation:

Given

- 6(p - 8) < - 12

Divide both sides by - 6, reversing the symbol as a result of dividing by a negative quantity.

p - 8 > 2 ( add 8 to both sides )

p > 10

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Please solve this it's on khan academy :(<br><br><br>Check Photo

Oduvanchick [21]

Answer:

y = 5x + 4

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 5 and c = 4, thus

y = 5x + 4 ← equation of line

6 0

3 years ago

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Which shows one way to determine the factors of x^3-12x^2-2x+24 by grouping

dezoksy [38]

Answer:

(x - 12)(x² - 2)

Step-by-step explanation:

Given

x³ - 12x² - 2x + 24 ( factor the first/second and third/fourth terms )

= x²(x - 12) - 2(x - 12) ← factor out (x - 12) from each term

= (x - 12)(x² - 2)

7 0

3 years ago

Sally needs to find the circumference of a circle with a diameter of 20 inches. She uses a calculator to find the circumference.

Vlad1618 [11]

Circumference = pi*diameter

C = pi*20

C = 20pi

When I used my calculator to compute 20*pi, I get approximately 62.83185

This is somewhat close to what choice C is saying. The difference is that choice C has an extra 1 buried inside it. I'm not sure if your teacher made a typo.

7 0

3 years ago

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Local versus absolute extrema. If you recall from single-variable calculus (calculus I), if a function has only one critical poi

stich3 [128]

Answer:

Step-by-step explanation:

Given that:

$f(x,y) = e^{3x} + y^3 - 3ye^x \\ \\ \implies \dfrac{\partial f}{\partial x} = 0 = 3e^{3x} -3y e^x = 0 \\ \\ e^{2x}= y \\ \\ \\ \implies \dfrac{\partial f}{\partial y } = 0 = 3y^2 -3e^x = 0 \\ \\ y^2 = e^x$

$\text{Now; to determine the critical point:}$ - $f_x = 0 ; \ \ \ \ \ f_y =0$

$\implies e^{2x} = y^4 = y \\ \\ \implies y = 0 \& y =1 \\ \\ since y \ne 0 , \ \ y = 1, \ \ x= 0\\\text{Hence, the only possible critical point= }(0,1)$

$\delta = f_xx, s = f_{xy}, t = f_{yy} \\ \\ . \ \ \ \ \ \ \ \ D = rt-s^2 \\ \\ i) Suppose D >0 ,\ \ \ r> 0 \ \text{then f is minima} \\ \\ ii) Suppose \ D >0 ,\ \ \ r< 0 \ \text{then f is mixima} \\ \\ iii) \text{Suppose D} < 0 \text{, then f is a saddle point} \\ \\ iv) Suppose \ D = 0 \ \ No \ conclusion$

$Thus \ at (0,1) \\ \\ \delta = f_{xx} = ge^{3x}\implies \delta (0,1) = 6 \\ \\ S = f_{xy} = -3e^x \\ \\ \implies S_{(0,1)} = -3 \\ \\ t = f_{yy} = 6y \\ \\$

$\implies t_{0,1} = 6$

$Now; D = rt - s^2 \\ \\ = (6)(6) -(-3)^2$

$= 36 - 9 \\ \\ = 27 > 0 \\ \\ r>0$

$\text{Hence, the critical point} \ (0,1) \ \text{appears to be the local minima}$

$\text{Suppose we chose x = 0 and y = -3.4} \\ \\ \text{Then, we have:} \\ \\ f(0,-3.4) = 1+ (-3.4)^3 + 3(3.4) \\ \\ = -28.104 < -1$

$\text{However, if f (0,1) = 1 +1 -3 = -1 \\ \\ f(0,-3.4) = -28.104} < -1} \\ \\ \text{This explains that} -1 \text{is not an absolute minimum value of f(x,y)}$

7 0

3 years ago

Mr. Abernathy bought a selection of wrenches for his shop and paid $78. He bought the same number of $1.50 and $2.50 wrenches, a

Dmitriy789 [7]

Let the number of $1.50 wrenches be x,
number of $2.50 wrenches = x
number of $4 wrenches = x/2
number of $3 wrenches = x/2 + 1

amount paid for $1.50 wrenches = 1.5x
amount paid for $2.50 wrenches = 2.5x
amount paid for $4 wrenches = 4(x/2)
= 2x
amount paid for $3 wrenches = 3(x/2 + 1)
= 1.5x + 3

total amount paid = 1.5x + 2.5x + 2x + 1.5x + 3
= 7.5x + 3
7.5x + 3 = 78
7.5x = 75
x = 10

number of $1.50 wrenches = x
= 10
number of $2.50 wrenches = x
= 10
number of $4 wrenches = x/2
= 10/2
= 5
number of $3 wrenches = x/2 + 1
= 10/2 + 1
= 5 + 1
= 6

He bought 10 $1.50 wrenches, 10 $2.50 wrenches, 5 $4 wrenches and 6 $3 wrenches.

7 0

3 years ago

Read 2 more answers