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2 years ago

He solution is n = –2 verified as a solution to the equation 1.4n + 2 = 2n + 3.2. What is the last line of the justification?

Mathematics

1 answer:

Nimfa-mama [501]2 years ago

5 0

Answer:

$-0.8=-0.8$

Step-by-step explanation:

We have been given that n=-2 verified as a solution to the equation $1.4n+2=2n+3.2$ .

As n=-2 is the solution for our given equation, so it will satisfy our equation.

Let us substitute n=-2 is our given equation.

$1.4*-2+2=2*-2+3.2$

$-2.8+2=-4+3.2$

$-0.8=-0.8$

Since the value of n satisfies our equation, therefore, last line of the justification will be $-0.8=-0.8$ .

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wolverine [178]

Answer:

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3 years ago

Alex skipped a jump rope 50 times every minute as shown by this table. x equals the number of minutes, and y equals the

KIM [24]

Answer:

Y = 50x

Since every minute he jumps 50 times, we multiply 50 by the amount of minutes (x)

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2 years ago

Which classification describes AMNO with vertices M(2, -3), N(3, 1), and<br> 0(-3, 1)?

Sonja [21]

Answer:

Option D

Step-by-step explanation:

32). Given vertices of the triangle are M(2, -3), N(3, 1) and O(-3. 1).

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the expression,

Distance = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between M(2, -3) and N(3, 1) will be,

MN = $\sqrt{(3-2)^2+(1+3)^2}$

= $\sqrt{1+16}$

= $\sqrt{17}$

Distance between M(2, -3) and O(-3, 1),

MO = $\sqrt{(2+3)^2+(-3-1)^2}$

= $\sqrt{25+16}$

= $\sqrt{41}$

Distance between N(3, 1) and O(-3, 1),

NO = $\sqrt{(3+3)^2+(1-1)^2}$

= 6

Condition for right triangle,

c² = a² + b² [Here c is the longest side of the triangle]

By this property,

MO² = MN² + NO²

$(\sqrt{41})^2=(\sqrt{17})^2+6^2$

41 = 17 + 36

41 = 51

False.

Therefore, given triangle is not a right triangle.

Since, length of all sides are not equal, given triangle will be a scalene triangle.

Option D is the correct option.

5 0

3 years ago

Nate works at least 4 hours but not more than 9 hours. He earns $12.40 per hour. The function f(t) = 12.4t represents the amount

Mashcka [7]

The practical domain is all real numbers from 4 to 9, inclusive.

The practical range is all real numbers from 49.6 to 111.6, inclusive.

These are the correct options.

Explanation:

Given function is f(t) = 12.4t

Let us assume that Nate works 'x' hours so 4<x<9

And multiplying the hours with his earnings we get the range.

4*12.40=49.6 and 9*12.40=111.6. Let the amount earned be represented by y

Hence, domain can be represented as 4<x<9 and range can be represented as 49.6 < y < 111.6

7 0

3 years ago