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nalin [4]
3 years ago
12

Solve x - 9< -5 A. X<4 B. X>-14 C. X>4 D. X<-14

Mathematics
2 answers:
Nookie1986 [14]3 years ago
8 0

Answer:

x<4

Step-by-step explanation:

x-9<-5

x<-5+9

x<9-5

x<4

Alchen [17]3 years ago
5 0

Answer:

A. X<4

Step-by-step explanation:

move the constant to the right hand side and change its sign

X<-5+9

calculate the sum

-5+9=4

X<4

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Vsevolod [243]

Answer:

12350%

Step-by-step explanation:

. 76 - 34 = 42

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. 1.23529411765 multiplied by 100 = 123.529411765

. 123.529411765 ≈ 123.5 * 100 = 12350%

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Evaluate the expression for n=-3
satela [25.4K]

Answer:

5n is your answer hope this help

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3 years ago
Verify the identity. 4 csc 2x = 2 csc2x tan x
vlada-n [284]

Step-by-step explanation:

4csc(2x) = 2csc^2(x) tan(x)

We start with Left hand side

We know that csc(x) = 1/ sin(x)

So csc(2x) is replaced by 1/sin(2x)

4 \frac{1}{sin(2x)}

Also we use identity

sin(2x) = 2 sin(x) cos(x)

4 \frac{1}{2sin(x)cos(x)}

4 divide by 2 is 2

Now we multiply top and bottom by sin(x) because we need tan(x) in our answer

2\frac{1*sin(x)}{sin(x)cos(x)*sin(x)}

2\frac{sin(x)}{sin^2(x)cos(x)}

2\frac{1}{sin^2(x)} \frac{sin(x)}{cos(x)}

We know that sinx/ cosx = tan(x)

Also  1/ sin(x)= csc(x)

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6 0
3 years ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
3 years ago
The standard tip in a restaurant is 15% of the bill before tax. Many people in California find the tip by doubling the sales tax
cricket20 [7]
To answer this you could do a couple things. 1. Actually calculate both tips and then subtract them. Here is the math ( assuming it is $60, not 60% that you mean). a. 0.15 x 60 = $9. b. 0.0825 x 2 x 60=$9.90. c. 9.90-9.00= $0.90 difference.

The other strategy would be to find the difference on the percents and then multiply by $60. a. 8.25%x2=16.5%-15%. Difference of 1.5%. b. 0.015 x $60=$0.90.
3 0
4 years ago
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