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dsp73
3 years ago
12

17.003 - 0.374 please show work

Mathematics
1 answer:
asambeis [7]3 years ago
7 0

Answer:

16.629

Step-by-step explanation:

Start by setting up the numbers like the first picture. The borrow. To borrow we have to go the whole way over to the 7 since you can't borrow from 0. So the 7 becomes a 6 and the 0 becomes  10. Then the 10 becomes a 9 and the 0 becomes a 10. Then the 10 becomes a 9 and the 3 becomes a 13. Then subtract. Just bring the decimal point down.

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Find the slope of the line through each pair of points. <br> (19, −16), (−7, −15)
erik [133]

Point-Slope:

y+16=-1/26*(x-19)

Finding the regular slope:

m=1/-26=0.03846

7 0
3 years ago
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alexgriva [62]
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4 0
4 years ago
10x^3 y^-5 z^-2 if x=3 y=2 and z=5 express your answer as a reduced fraction
sineoko [7]

Answer:

\large\boxed{\dfrac{27}{80}}

Step-by-step explanation:

Put x = 3, y = 2 and z = 5 to the given expression 10x^3y^{-5}z^{-2}:

10(3^3)(2^{-5})(5^{-2})\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{10(27)}{(2^5)(5^2)}=\dfrac{270}{(32)(25)}=\dfrac{270}{800}=\dfrac{27}{80}

7 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
Which real numbers are zeros of the function?
monitta
<h3><u>The roots are -1/2, 0, 2, and 3.</u></h3>

Let's trying factoring this polynomial.

We can factor an x out of each term to start.

x(2x^3 - 9x^2 + 7x + 6)

We now know one of the roots is going to be zero.

Using the rational roots theorem, and the remainder theorem, we can try to find some more roots that way.

Factors of 6: 1, 2, 3, 6.

Factors of 2: 1

Possible rational roots: +/-1/2, +/-1, +/-2, +/-3/2, +/-3, +/-6

Using the remainder theorem, we can plug these values into the polynomial, and if we get a remainder of zero, we know it's a root.

-1/2(2(-1/2)^3 - 9(-1/2)^2 + 7(-1/2) + 6) = 0

Our first root is -1/2.

We can successfully factor a -1/2 out of this polynomial.

After diving the polynomial by -1/2, we're left with: 2x^2 - 10x + 12.

We can now try using the AC method to get our last two roots.

First, we can factor this polynomial to simplify it.

Divide all terms by 2.

2(x^2 - 5x + 6)

Now let's try using the AC method.

The digits -3 and -2 satisfy the criteria.

2(x - 3)(x - 2)

We now have all of our roots: 0, -1/2, 2, and 3.


6 0
4 years ago
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