The first parenthesis is $13z+8$ . These are not like terms because one involves a variable, while the other is a constant term. Two terms are summable if they involve the same power(s) of the same variable(s).

So, we can take one step outwards, and we arrive to the square brackets. We have

$2z-13z-8$

and 2z and -13z are like terms, so we can sum them:

$2z-(13z +8) = 2z-13z-8 = (2-13)z-8 = -11z-8$

Finally we arrive to the whole expression, which is

$5z-(-11z-8) = 5z+11z+8 = (5+11)z+8=16z+8$

Because, again, 5z and 11z were like terms.

6 0

3 years ago

Write a word problem for the inequality -3.5 < -2.5

nlexa [21]

Jack is in debt after borrowing some money. He has more than -3.5 dollars and less than -2.5 dollars. Write this as an inequality.

8 0

2 years ago

If measure of angle 1 = 2x+6 and measure of angle 2 = 8x+4 where angles 1 and 2 are supplementary, then how much is the actual m

nlexa [21]

Answer:

The actual measure of ∠1 = 40°

Step-by-step explanation:

We are given: ∠1 = 2x+6 and ∠2 = 8x+4

and both angles are supplementary. we need to find the actual measure of ∠1

Supplementary angles: If the angles are supplementary their sum is equal to 180°

In our case ∠1 +∠2 = 180°

2x+6+8x+4=180

Solving the equation to find x, then we can find the actual measure of ∠1

$2x+6+8x+4=180\\10x+10=180\\10x=180-10\\10x=170\\x=170/10\\x=17$

So, value of x = 17

Now, finding actual measure of ∠1 by putting x=17 in equation 2x+6

$\angle1=2(17)+6\\\angle1=34+6\\\angle1=40$

So, the actual measure of ∠1 = 40°

8 0

2 years ago

A Find tan a. r √5,-√7) ✓[? Enter

Virty [35]

Answer:

$tan(\alpha)=-\frac{\sqrt{35}}{5}$

Step-by-step explanation:

Tan can be defined as: $\frac{sin(\theta)}{cos(\theta)}$ as it simplifies to opposite/adjacent. If you know a bit about the unit circle, you'll know that the x-coordinate is going to be cos(theta) and the y-coordinate is going to be sin(theta). Since the sin(theta) is defined as opposite/hypotenuse, and the hypotenuse = 1, so sin(theta) is defined as the opposite side, which is the y-axis. Same thing goes for cos(theta), except the adjacent side is the x-axis.

Using this we can define tan

$sin(\alpha)=-\sqrt{7}\\cos(\alpha)=\sqrt{5}\\\\tan(\alpha)=-\frac{\sqrt{7}}{\sqrt{5}} * \frac{\sqrt{5}}{\sqrt{5}}\\tan(\alpha)=-\frac{\sqrt{7*5}}{5}\\tan(\alpha)=-\frac{\sqrt{35}}{5}\\$

8 0

2 years ago

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