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Sati [7]
3 years ago
11

Solve the following equation 6x-1+7x=2x+2-4x

Mathematics
2 answers:
Ipatiy [6.2K]3 years ago
7 0

Answer:

x=5 hi hoped this helped

ehidna [41]3 years ago
5 0

Answer:The result can be shown in multiple forms.

Exact Form:

x = 1/ 5

Decimal Form:

x = 0.2

Step-by-step explanation:

6 x − 1 + 7 x = 2 x + 2 − 4 x

Add  6 x  and  7 x .

13 x − 1 = 2x + 2 − 4 x

Subtract  4 x  from  2 x .

13 x − 1 = − 2 x + 2

Move all terms containing  x  to the left side of the equation.

15 x − 1 = 2

Move all terms not containing  x  to the right side of the equation.

15 x = 3

Divide each term by  15  and simplify.

x = 1 /5

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How many solutions does the equation have? 2 + 34x − 5 = 3x − 3 − 94x A. 0 B. 1 C. 2 D. infinite
Molodets [167]

Answer:

0

Step-by-step explanation:

Here, we want to find the number of solutions that the equation have.

We can easily find this by solving the equation;

2 + 34x -5 = 3x -3-94x

2-5 + 34x = 3x -94x -3

-3 + 34x = -91x -3

Now since we have -3 on both sides, we can easily cancel out and we are left with;

34x = -91x

This shows that we cannot find x and thus the equation has no solution

4 0
4 years ago
There is a quadrilateral MNPQ in which side MN is congruent to side PQ and side NP is parallel to side MQ. The diagonal MP and t
8_murik_8 [283]

Answer: QN = 12

Step-by-step explanation: This quadrilateral is a paralelogram because its 2 opposite sides (NP and MQ) are parallel and the other 2 (MN and PQ) are congruent.

In paralelogram, diagonals bisect each other, which means QR = RN.

If QR = RN:

QR = 6

Then,

QN = QR + RN

QN = 6 + 6

QN = 12

<u>The diagonal QN of quadrilateral MNPQ is </u><u>QN = 12</u><u>.</u>

5 0
3 years ago
Plzz help only 10 minutes left to finish
Svet_ta [14]
I’m really sorry I’m not sure
5 0
3 years ago
The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s. At
musickatia [10]

Answer:

Volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

Step-by-step explanation:

Given: The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s.

To find: The rate at which volume of the cone changing when the radius is 134 in. and the height is 136 in.

Solution:

We have,

\frac{dr}{dt} =1.9 \:\text{in/s}, \frac{dh}{dt}=-2.2\:\text{in/s}, r=134 \:\text{in}, h=136\:\text{in}

Now, let V be the volume of the cone.

So, V=\frac{1}{3}\pi r^{2}h

Differentiate with respect to t.

\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ]

Now, on substituting the values, we get

\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left (  136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]

\frac{dv}{dt} =\frac{1}{3}\pi\left [  -39503.2+69251.2 \right ]  

\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]

\frac{dv}{dt} =9916\pi \frac{in^3}{s}

Hence, the volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

6 0
3 years ago
Did I put the correct answer? Please let me know ASAP! See the attachment.
Nataliya [291]
Yes the answer is correct
8 0
3 years ago
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