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lisov135 [29]
3 years ago
13

10=4x-2 what do i ????

Mathematics
2 answers:
laiz [17]3 years ago
5 0

Hey there!

x = 3 \\ 10 = 4(3)  -2  \\ 4(3) = 12  \\12 - 2 = 10 \\  10 = 10  \\ \\ Answer: x = 3

<h2>Good luck on your assignment and enjoy your day! </h2>

~LoveYourselfFirst:)

Finger [1]3 years ago
5 0

First, add 2 to both sides: 10(+2)=4x-2(+2) Making the 2 on the right cancel out and the 10 goes up to 12.

Now all you should have is 12=4x, so divide both sides by 4, both 4's cancel out and then you're left with 3 (since 12 divided by 4 is 3).

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Norma-Jean [14]

Answer:

2.01 x 15 = 30.15

Step-by-step explanation:

2.01 x 15 = 30.15

or

201 * .15 = 30.15

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3 years ago
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Which polynomials are listed with their correct additive inverse? Check all that apply. x2 + 3x – 2; –x2 – 3x + 2 –y7 – 10; –y7
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a c d. simple

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How to work this out ?
Free_Kalibri [48]

Answer:

The solution is (-3, -2).

Step-by-step explanation:

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6 0
2 years ago
What is the distance between the points (-7, 5) and (-2, -7)
harkovskaia [24]

Answer:

The distance between those points is 13.

Step-by-step explanation:

1.    (x2-x1) = (-2 - -7) = 5

2.   (y2-y1) = (-7 - 5) = -12

3.   (5)2 + (-12)2 = 25 + 144 = 169

4.   √169 = 13

5.   13

7 0
3 years ago
The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?
omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

T_5 = 160

T_7 = 40

Required

T_6

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

\frac{T_7}{T_5} = \frac{40}{160}

Divide the numerator and the denominator of the fraction by 40

\frac{T_7}{T_5} = \frac{1}{4} ----- equation 1

Recall that the formula of a GP is

T_n = a r^{n-1}

Where n is the nth term

So,

T_7 = a r^{6}

T_5 = a r^{4}

Substitute the above expression in equation 1

\frac{T_7}{T_5} = \frac{1}{4}  becomes

\frac{ar^6}{ar^4} = \frac{1}{4}

r^2 = \frac{1}{4}

Square root both sides

r = \sqrt{\frac{1}{4}}

r = ±\frac{1}{2}

Next, is to solve for the first term;

Using T_5 = a r^{4}

By substituting 160 for T5 and ±\frac{1}{2} for r;

We get

160 = a \frac{1}{2}^{4}

160 = a \frac{1}{16}

Multiply through by 16

16 * 160 = a \frac{1}{16} * 16

16 * 160 = a

2560 = a

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

T_n = a r^{n-1}

Here, n = 6;

T_6 = a r^{6-1}

T_6 = a r^5

T_6 = 2560 r^5

r = ±\frac{1}{2}

So,

T_6 = 2560( \frac{1}{2}^5) or T_6 = 2560( \frac{-1}{2}^5)

T_6 = 2560( \frac{1}{32}) or T_6 = 2560( \frac{-1}{32})

T_6 = 80 or T_6 = -80

T_6 =±80

Hence, the 6th term is positive/negative 80

8 0
2 years ago
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