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Andru [333]
3 years ago
15

Solve by using subitution

Mathematics
2 answers:
zhenek [66]3 years ago
5 0

Answer:

No Solutions

Step-by-step explanation:

1. Substitute the 2x + 1 in for the y in the first equation. Rewrite it as 2x - (2x + 1) = 2.

2. Add the opposite and change the equation to 2x + -1(2x + 1) = 2.

3. Use the distributive to make the equation look like this: 2x + -2x + -1 = 2

4. Combine like terms and add 2x and -2x. They cancel each other out.

5. You will end up with -1 ≠ 2. Since -1 and 2 are not equal, there are no solutions.

Bas_tet [7]3 years ago
4 0

Y=2x+1
2x-y=2
2x-(2x+1)=2
-1=2
-1+1=2+1
0=3


No solution
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Find the volume of a pyramid with a square base, where the side length of the base is 10.9 m and the height of the pyramid is 4.
natita [175]

Answer:

V=186.1\ m^3

Step-by-step explanation:

<u>Volume of a Pyramid</u>

The volume of a pyramid with base area Ab and height H is given by:

\displaystyle V=\frac{A_b*H}{3}

If the base is a square of side length L, then its area is:

A_b=L^2

Thus, substituting the above equation in the first:

\displaystyle V=\frac{L^2*H}{3}

The side length is L=10.9 m and the height is 4.7 m, thus the volume is:

\displaystyle V=\frac{10.9^2*4.7}{3}

\displaystyle V=\frac{558.407}{3}

\boxed{V=186.1\ m^3}

3 0
4 years ago
We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance
MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

h(t) = -16\cdot t^{2}+256\cdot t+1680 (1)

Where:

h - Height above the ground, measured in feet.

t - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680

h(3\,s) = 2304\,ft

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at t = 5\,s and t = 7\,s:

h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680

h(5\,s) = 2560\,s

h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680

h(7\,s) = 2688\,s

We notice that the small object ascends in the given interval.

\Delta h = h(7\,s)-h(5\,s)

\Delta h = 128\,ft

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that h = 2640\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t-960=0 (2)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 10\,s and t_{2}= 6\,s

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that h = 0\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t +1680 = 0 (3)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 21\,s and t_{2} = -5\,s

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

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Try dividing 47 by 4.75:)

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Answer:

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Step-by-step explanation:

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