(6.8*10^4) +(9.24*10^5)

1) What can u do to get 2x alone on the left side?

CaHeK987 [17]

Answer:

1)

82.5 - 82.5 + 2x = 338.5 - 82.5

2x = 256

2)

2x ÷ 2 = 256 ÷ 2

x = 128

Step-by-step explanation:

1)

In order to get rid 82.5, you have to substract the same value on both side to get a 0 value.

2)

In order to find the value of x, you have to divide the value that is sticked with x which is 2. As you divide 2 by 2, you will get 1 as a single value of x so you have to divide 2 to both side too.

5 0

2 years ago

Find equations for the lines through the point (a,

Free_Kalibri [48]

The line y = mx + c has a slope of m.

Parallel line:
A parallel line will also have a slope of m, and should be of the form
y = mx + k
Because the line passes through the point (a,b), therefore
ma + k = b
k = b - ma
The equation of a parallel line is
y = mx + b - ma
= m(x-a) + b

Perpendicular line:
A perpendicular line will have a slope of -1/m, because the product of slopes should be -1.
The equation for a perpendicular line is of the form
y = -(1/m)x + k
Because the line passes through (a,b), therefore
-(1/m)a + k = b
k = b + a/m
The equation for a perpendicular line is
y = -(x/m) + b + a/m
= -(1/m)(x-a) + b

Answer:
Parallel line: $y=m(x-a)+b$
Perpendicular line: $y=- \frac{1}{m}(x-a)+b$

3 0

3 years ago

For which of the following compound inequalities is there no solution?

Bess [88]

Answer:

(a) $6m \le - 36$ and $m + 24 > 20$

Step-by-step explanation:

Required

Which of a to d has no solution

(a) $6m \le - 36$ and $m + 24 > 20$

We have: $m + 24 > 20$

Sole for m

$m >20 - 24$

$m >-4$

$6m \le - 36$

Divide by 6

$m \le -6$

So, we have:

$m \le -6$ and $m >-4$

$m \le -6$ implies that: m = -6,-7,-8,-9.....

$m >-4$ implies that m = -3,-2,-1,0.....

Hence, there is no solution

8 0

2 years ago

Calculate limits x>-infinity -2x^5-3x+1

Lera25 [3.4K]

Given:

The limit problem is:

$\lim_{x\to -\infty}(-2x^5-3x+1)$

To find:

The value of the given limit problem.

Solution:

We have,

$\lim_{x\to -\infty}(-2x^5-3x+1)$

In the function $-2x^5-3x+1$ , the degree of the polynomial is 5, which is an odd number and the leading coefficient is -2, which is a negative number.

So, the function approaches to positive infinity as x approaches to negative infinity.

$\lim_{x\to -\infty}(-2x^5-3x+1)=\infty$

Therefore, $\lim_{x\to -\infty}(-2x^5-3x+1)=\infty$ .

3 0

2 years ago

What is the sum (Two-fifths x + StartFraction 5 over 8 EndFraction) + (one-fifth x minus one-fourth)?

liubo4ka [24]

The sum of ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) is $\frac{3}{5}$ x + $\frac{3}{8}$

Step-by-step explanation:

To add or subtract 2 fractions, they must have same denominators, if they not do that

Find the Lowest common multiple of the two denominators (LCM)
Replace each denominator by it
Divide The LCM by each denominator and multiply the numerator of each fraction by its quotient

∵ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ )

- Let us start with adding the like terms

∴ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) = (

∵ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) have same denominators, then we can add them

∴ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) = $\frac{3}{5}$ x

∵ ( $\frac{5}{8}$ - $\frac{1}{4}$ ) do not have the same denominators, then we must find

LCM of 8 and 4

- The LCM of 8 and 4 is 8 because 8 is the first common multiple

of 8 and 4

∵ LCM of 8 and 4 is 8

- Divide 8 by the denominator 4

∵ 8 ÷ 4 = 2

- Multiply the numerator of the fraction by 2

∴ $\frac{1}{4}=\frac{2}{8}$

∴ ( $\frac{5}{8}$ - $\frac{1}{4}$ ) = ( $\frac{5}{8}$ - $\frac{2}{8}$ ) = $\frac{3}{8}$

∴ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) = $\frac{3}{5}$ x + $\frac{3}{8}$

The sum of ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) is $\frac{3}{5}$ x + $\frac{3}{8}$

Learn more:

You can learn more about the fractions in brainly.com/question/2456302

#LearnwithBrainly

7 0

2 years ago

Read 2 more answers