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3 years ago

Helpppppppppppppppppp

Mathematics

2 answers:

Vesnalui [34]3 years ago

6 0

The answer is on the bottom left

xxTIMURxx [149]3 years ago

4 0

Answer is C.

If its wrong im sorry but hope this helps.

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C(x)=x^2+6x+14 find vertex form

nalin [4]

Answer:

c(x)=(x+3)^2+5

Step-by-step explanation:

To complete the square, the same value needs to be added to both sides.

So, to complete the square x^2+6x+9=(x+3)^2 add 9 to the expression

C(x) =x^2 +6x + 9 + 14

Since 9 was added to the right-hand side also add 9 to the left-hand side

C(x) +9= x^2 +6x + 9 + 14

Using a^2 + 2ab + b^2=(a+b)^2, factor the expression

C(x)+9= (x+3)^2 +14

Move constant to the right-hand side and change its sign

C(x)=(x+3)^2 +14 - 9

Subtract the numbers

C(x)= (x+3)^2 +5

6 0

3 years ago

I just started school and they hit my with this. -6s=35 Can someone help?

mihalych1998 [28]

Answer:

s = -35/6

Step-by-step explanation:

Let's evaluate your equation:

-6s = 35 <---- In order to evaluate this equation we must divide.

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

-6 -6

s = -35/6 <--- Final answer :D

I hope this helps :)

7 0

3 years ago

Read 2 more answers

Which number is greatest? O 2.89 x 10 O 1.997x102 O 8.9x104 O 5x10

seropon [69]

Answer:

Step-by-step explanation:

you just multiply in the calculator

8 0

3 years ago

Which portion of the unit circle satisfies the trigonometric inequality cos^2theta + sin^2theta is greater than or equal to 1. A

liberstina [14]

Answer:

Only points on the circle satisfy the given inequality.

Step-by-step explanation:

Given: Unit circle

To find: portion of the unit circle which satisfies the trigonometric inequality $\sin ^2\theta +\cos ^2\theta \geq 1$

Solution:

In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)

$\sin \theta$ = side opposite to $\theta$ /hypotenuse

$\cos \theta$ = side adjacent to $\theta$ /hypotenuse

$\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta$

$\cos \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta$

So, coordinates of A = $\left ( \cos \theta ,\sin \theta \right )$

For any point (x,y) on the unit circle with centre at origin, equation of circle is given by $x^2+y^2=1$

Put $(x,y)=\left ( \cos \theta ,\sin \theta \right )$

$\cos ^2\theta +\sin ^2\theta =1$

So, $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ satisfies the equation $x^2+y^2=1$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ inside the circle, $\cos ^2\theta +\sin ^2\theta$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ outside the circle, $\cos ^2\theta +\sin ^2\theta >1$

So, only points on the circle satisfy the given inequality.

4 0

3 years ago

Read 2 more answers

Click once to choose an answer. Click again to change your answer. Choose all that locate the ordered pair in the correct quadra

forsale [732]

Answer:

$Quadrant\ IV = (5, -3)$

$Quadrant\ I = (9, 11)$

$Quadrant\ II = (-1, 3)$

$Quadrant\ III = (-2, -4)$

Step-by-step explanation:

Given

$Quadrant\ IV = (5, -3)$

$Quadrant\ II = (-4, -8)$

$Quadrant\ I = (9, 11)$

$Quadrant\ II = (-1, 3)$

$Quadrant\ III = (-2, -4)$

Required

Determine which coordinate fall in the right quadrant

First, we split the each quadrant into x and y axis

In the first:

x and y is +

In the second:

x is - and y is +

In the third

x and y are -

In the fourth

x is + and y is -

Comparing the given coordinates to their respective quadrants, base on the conditions stated above

$Quadrant\ IV = (5, -3)$ is correct

$Quadrant\ I = (9, 11)$ is correct

$Quadrant\ II = (-1, 3)$ is correct

$Quadrant\ III = (-2, -4)$ is correct

$Quadrant\ II = (-4, -8)$ is incorrect

3 0

3 years ago