Does anyone know this to help me out

Answer:

18

Step-by-step explanation:

6+3=9

9*2=18

sorry if not right

Answer:

hope this helps

Step-by-step explanation:

Answer:

neither

Step-by-step explanation:

they dont intersect with each other

Answer: Its 516. Hope this helps!

Step-by-step explanation:

Answer:

516

Step-by-step explanation:

Answer:

b. angles 1 and 8

Step-by-step explanation:

alternate exterior angles theorm

Answer:

92

Step-by-step explanation:

If the 4 consecutive even integers have a sum of 380, then the average value is 380/4.

380/4 = 95

There must be 2 integers less than 95 and 2 grater than 95.

The 4 integers are: 92, 94, 96, 98

Sum: 92 + 94 + 96 + 98 = 380

Answer: 92

The 4 consecutive even integers are 92,94,96 and 98 whose sum is 380.

Sequence is an enumerated collection of objects in which repetitions are allowed and order matters.

The  4 consecutive even integers has a sum of 380.

x+2+x+4+x+6+x+8=380

Add the like terms

4x+20=380

Subtract 20 on both sides

4x=380-20

4x=360

Divide both sides by 4

x=90

So the four consecutive integers are

x+2=90+2=92

x+4=90+4=94

x+6=90+6=96

x+8=90+8=98

Hence, the 4 consecutive even integers are 92,94,96 and 98 whose sum is 380.

To learn more on Sequence click:

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Answer:

11:30 AM

Step-by-step explanation:

Answer:

55

Step-by-step explanation:

No matter what happens in between, the average rate of change between 2 points of any interval is the slope of the 2 points of the interval.

The 2 points are (-8, -50) and (-6, 60), and we need to find the slope (which, as a reminder, is change in x over change in y):

(60-(-50))/(-6-(-8)) = (110)/(2)=55

Therefore, the average rate of change between the interval -8≤x≤-6 is 55.

Answer:

y = 2/3x +6  for x< -3

y = 2/3x +1     for x> 3

Step-by-step explanation:

The graph is a line for  x < -3

( - 6,2) and ( -3,4)

The slope is m= ( y2-y1)/(x2-x1)

m = ( 4-2)/(-3 - -6) = 2/ ( -3+6) = 2/3

The slope is 2/3

Using point slope form

y-y1 = m(x-x1)  and the point ( -6,2)

y -2 = 2/3(x - -6)

y -2 = 2/3(x +6)

y-2 = 2/3 x +4

y = 2/3x +6 x< -3

The graph is a line for  x > 3

(3,3) and ( 6,5)

The slope is m= ( y2-y1)/(x2-x1)

m = ( 5-3)/(6-3) = 2/ (3) = 2/3

The slope is 2/3

Using point slope form

y-y1 = m(x-x1)  and the point (6,5)

y -5 = 2/3(x - 6)

y-5 = 2/3 x -4

y = 2/3x +1   x> 3

Answer:

x= -5

Step-by-step explanation:

just graphed it . y = 4 is a flat horizontal line.

x=5 is a vertical line that passes through (5,2)

Answer:

3 2/3

Step-by-step explanation:

divide the 3 by 11 and your product is the whole number, with the remainder being the numerator.

11 / 3 = 9.2

9/3=3

3 2/3

Answer:

It is not true

Step-by-step explanation:

-4 is closer to 0 than -5 1/2

The closer the negative number to 0 the bigger it is

Answer:

62

Step-by-step explanation:

62

Answer:

x=40/21

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−3x−3(6x−11)=−7

−3x+(−3)(6x)+(−3)(−11)=−7(Distribute)

−3x+−18x+33=−7

(−3x+−18x)+(33)=−7(Combine Like Terms)

−21x+33=−7

−21x+33=−7

Step 2: Subtract 33 from both sides.

−21x+33−33=−7−33

−21x=−40

Step 3: Divide both sides by -21.

−21x

−21

=

−40

−21

Answer:

9

Step-by-step explanation:

f(8)= 1/2(8) + 5

f(x)= 4 +5

f(x)= 9

Unfortunately sucrose isn't classified as any of the above because it is not alive. Sucrose can however be classified as a disaccharide because it contains two monosaccharides (glucose and fructose). If you want to be more broad you can classify it as a carbohydrate.

Answer:  see proof below

Step-by-step explanation:

Use the following Half-Angle Identities:

sin² A = (1 - cos 2A)/2

cos² A = (1 + cos 2A)/2

Proof LHS → RHS:

LHS:                          sin⁴ A

Expand:                     sin² A · sin² A

[tex]\text{Half-Angle:}\qquad \qquad \bigg(\dfrac{1-\cos (2A)}{2}\bigg)\bigg(\dfrac{1-\cos (2A)}{2}\bigg)[/tex]

[tex]\text{Distribute:}\qquad \qquad \dfrac{1-2\cos (2A)+\cos^2 (2A)}{4}[/tex]

[tex]\text{Half-Angle:}\qquad \qquad \dfrac{1-2\cos (2A)+\frac{1+\cos (2\cdot 2A)}{2}}{4}[/tex]

[tex]\text{Simplify:}\qquad \qquad \dfrac{\frac{2}{2}[1-2\cos (2A)]+\frac{1+\cos (4A)}{2}}{4}[/tex]

                         [tex]=\dfrac{2-4\cos (2A) + 1 + \cos (4A)}{8}[/tex]

                         [tex]=\dfrac{1}{8}\bigg(3-4\cos (2A)+\cos (4A)\bigg)[/tex]

LHS = RHS  [tex]\checkmark[/tex]

Let's start from LHS...

LHS = sin^4A = sin^2A * sin^2A

We can apply a few formulas here;

cos 2A = 1 - 2sin^2A

sin^2A = (1 - cos2A) / 2

cos2A = 2cos^2A - 1

cos^2A = (1 + cos2A) / 2

According to the second formula, sin^2A * sin^2A = (1 - cos2A) / 2 * (1 - cos2A) / 2.

(1 - cos2A) / 2 * (1 - cos2A) / 2 = 1/4(1 - cos2A)^2

We can simplify "(1 - cos2A)^2." Remember that (a - b)^2 = a^2 - 2ab + b^2. Similarly (1 - cos2A)^2 = (1 - 2cos 2A + cos^2 2A);

1/4(1 - cos2A)^2 = 1/4(1 - 2cos 2A + cos^2 2A)

According to the fourth formula, cos^2 2A = (1 + cos4A) / 2;

1/4(1 - 2cos 2A + cos^2 2A) = 1/4(1 - 2cos 2A + (1 + cos4A) / 2)

= (taking LCM) 1/4( (2 - 4cos 2A + 1 + cos4A)/2 )

= (simplified) 1/8(3 - 4cos 2A + cos 4A)

Answer:

C.

Step-by-step explanation:

Answer:

George's mistake would be:

Instead of 2x + 40, she has made - x + 20.

Step-by-step explanation:

Slope - Intercept Form = y = mx + b

y + 20 = 2(x +20)

y + 20 = 2x + 40

   - 20          - 20

----------------------

y = 2x + 20

(Should have been the answer)

BUT,

George's Solving:

y + 20 = 2(x + 20)

y + 20 = -x + 20

George's Mistake would be her miscalculation during her 2nd step. Where instead of distributing 2 to (x + 20), she has changed it to -x + 20, when in reality, it should have been 2x + 40.

Answer:

the answer would be C

Step-by-step explanation:

the opposite of exponents is square root it

Answer:

A. c. F.

Step-by-step explanation:

Answer:

X=2

Step-by-step explanation:

Make an equation: the scales are equal, so

x + x + 3 = 7

2x + 3 = 7

2x = 4

x = 2

Answer:

enough is the correct worf

ANSWER:

8,7

5,6

4,3

1,2

This is a angle of pair

Consider the operation is [tex]R_2\to -3R_2[/tex].

Given:

The augmented matrix below represents a system of equations.

[tex]\left[\left.\begin{matrix}1&0&1\\1&3&-1\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-9\\-2\end{matrix}\right][/tex]

To find:

Matrix results from the operation [tex]R_2\to -3R_2[/tex].

Step-by-step explanation:

We have,

[tex]\left[\left.\begin{matrix}1&0&1\\1&3&-1\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-9\\-2\end{matrix}\right][/tex]

After applying [tex]R_2\to -3R_2[/tex], we get

[tex]\left[\left.\begin{matrix}1&0&1\\-3(1)&-3(3)&-3(-1)\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-3(-9)\\-2\end{matrix}\right][/tex]

[tex]\left[\left.\begin{matrix}1&0&1\\-3&-9&3\\3&2&0\end{matrix}\right|\begin{matrix}-1\\27\\-2\end{matrix}\right][/tex]

Therefore, the correct option is A.