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MAXImum [283]
1 year ago
15

Which interval describes where the graph of the function is positive?

Mathematics
2 answers:
Ad libitum [116K]1 year ago
8 0

Answer:

D is the correct answer.

Step-by-step explanation:

Find the portion of the graph that is above the x-axis.

Klio2033 [76]1 year ago
7 0

Answer:

D

Step-by-step explanation:

Coming in from - inf the graph is positive and remains so until yo get to x = -1  <==== then it is zero    and continues to negative values

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A small bar of gold measures 20 mm by 300 mm by 2 mm. One cubic millimeter of gold weighs about 0.0005 ounces. Find the volume i
Keith_Richards [23]

Answer:

12000 mm^2

6 ounces

Step-by-step explanation:

20*300*2=12000

12000*0.0005=6

4 0
3 years ago
Each leg of a 45-45-90 triangle has a length of 6 units. What is the length of its hypotenuse?
andreyandreev [35.5K]
Use the Pythagorean theorem:
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The length of the hypotenuse is 6√2 units.
5 0
3 years ago
Can you pls help? This is a multiple choice, and including the answer can u add a one sentence explanation on why? Tysm! I’ll gi
prohojiy [21]

Answer:

A and D

Step-by-step explanation:

5 0
2 years ago
20 is how much less than 30? What percent less?
LenKa [72]

Answer:

33.3333% decrease

Step-by-step explanation:

Percent change formula: \frac{(x-y)}{|y|} *100, where x is the new value and y is the starting value.

Plug in values and solve:

\frac{20-30}{30} *100\\\frac{-10}{30} *100\\-0.3333*100\\

-33.3333% change, or a 33.3333% decrease.

7 0
3 years ago
(2*)2-3×2*+2=0<br>4m-15°(×)m+75°​
Valentin [98]

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

7 0
2 years ago
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