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3 years ago

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

Mathematics

2 answers:

laiz [17]3 years ago

5 0

Hey there!

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

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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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Step-by-step explanation:

2.01 x 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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3 years ago

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How to work this out ?

Free_Kalibri [48]

Answer:

The solution is (-3, -2).

Step-by-step explanation:

Please note that if you simply add these 2 equations together, 9x - 9x = 0, and you are then left with -5y = 10. Thus, y = -2. Subbing -2 for y in the first equation, we get 9x - 8(-2) = -11, or 9x + 16 = -11.

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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

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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