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ANEK [815]
3 years ago
15

Solve for X-2x-4=-4x+3​

Mathematics
2 answers:
boyakko [2]3 years ago
6 0

Answer:

x=0

Step-by-step explanation:

Feliz [49]3 years ago
3 0

Answer:

Step-by-step explanation:Let's solve your equation step-by-step.

−

2

x

−

4

=

−

4

x

+

3

Step 1: Add 4x to both sides.

−

2

x

−

4

+

4

x

=

−

4

x

+

3

+

4

x

2

x

−

4

=

3

Step 2: Add 4 to both sides.

2

x

−

4

+

4

=

3

+

4

2

x

=

7

Step 3: Divide both sides by 2.

2

x

2

=

7

2

x

=

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ololo11 [35]

Their monthly income is 4,650.

Multiply that by 12 to find the yearly income:

4650 x 12 = $55,800


Based on the yearly income the answer is D. Virginia or Texas

7 0
3 years ago
Read 2 more answers
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.

8 0
3 years ago
can someone show me how this problem would work out? I dont just want an answer I want to see how this problem is worked out. Th
densk [106]

Answer:

No probllem linked

Step-by-step explanation:

3 0
3 years ago
0.6y+0.02=0.3y-0.4 value of y?
AleksandrR [38]

0.6y + 0.02 = 0.3y - 0.4

Subtract 0.3y from both sides.

0.3y + 0.02 = -0.4

Subtract 0.02 from both sides.

0.3y = -0.42

Divide both sides by 0.3.

y = -1.4

4 0
3 years ago
2.5.3 Quiz Inverse Functions Question 2 of 10 f (1) x+ 10. Find the inverse of (x)​
Vanyuwa [196]

Answer:

f^-1 (x) = x - 10...........

5 0
3 years ago
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