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3 years ago

15

PLEASE HELP ME!!!

1 answer:

Harman [31]3 years ago

8 0

Send a better photo or tell me what I’m suppose to be answering

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-6x-8<4 plz help me this is sixty percent of my grade

masya89 [10]

Answer:

x>-2

Step-by-step explanation:

5 0

3 years ago

Angle θ lies in the second quadrant, and sin θ = 3/5.<br> cos θ = <br><br> tan θ =

storchak [24]

Answer:

$cos\Theta =\frac{4}{5}$

$tan\Theta =\frac{3}{4}$

Step-by-step explanation:

Given : sin θ = 3/5

To Find : value of cos θ and tan θ

Solution :

use the identity:

$sin^{2}\Theta +cos^{2}\Theta =1$

putting value of sin θ

⇒ $(\frac{3}{5})^{2} + cos^{2}\Theta =1$

⇒ $\frac{9}{25} +cos^{2}\Theta =1$

⇒ $cos^{2}\Theta = 1-\frac{9}{25}$

⇒ $cos^{2}\Theta = \frac{16}{25}$

⇒ $cos\Theta = \sqrt{\frac{16}{25}}$

⇒ $cos\Theta =\frac{4}{5}$

Thus , $cos\Theta =\frac{4}{5}$

Now to find value of tan θ

Since we know that

⇒ $tan\Theta =\frac{sin\Theta }{cos\Theta }$ (identity)

⇒ $tan\Theta =\frac{\frac{3}{5} }{\frac{4}{5} }$

⇒ $tan\Theta =\frac{3}{5}\div \frac{4}{5}$

⇒ $tan\Theta =\frac{3}{5}\times \frac{5}{4}$

⇒ $tan\Theta =\frac{3}{4}$

Thus , the value of

$tan\Theta =\frac{3}{4}$

$cos\Theta =\frac{4}{5}$

3 0

3 years ago

Read 2 more answers

If y=3 when x=4, find y when x=6

rodikova [14]

Answer:

y = 2

Step-by-step explanation:

(x)(y) = k Inverse relationship

(4)(3) = k

12 = k

(6)(y) = 12

y = 2

7 0

3 years ago

Read 2 more answers

Find the measure of

Irina-Kira [14]

Answer:3Y+6=12

3(2)+6=12

Step-by-step explanation:Because 3 + 3 is 6 So That means that you repeat the pattern 4 times 3 Times 4 equals 12

4 0

3 years ago

Find all solutions to the equation. -\dfrac{2}{x(x+2)}=\dfrac{x+3}{x+2}− x(x+2) 2 = x+2 x+3 minus, start fraction, 2, divide

Gnoma [55]

Answer:

$x=-1$

Step-by-step explanation:

We have been given an equation $\frac{-2}{x(x+2)}=\frac{x+3}{x+2}$ . We are asked to find the solutions of our given equation.

First of all, we will cross multiply our given equation as:

$-2(x+2)=x(x+2)(x+3)$

Subtract $x(x+2)(x+3)$ from both sides:

$-2(x+2)-x(x+2)(x+3)=x(x+2)(x+3)-x(x+2)(x+3)$

$-2(x+2)-x(x+2)(x+3)=0$

Now we will factor out $(x+2)$ .

$(x+2)(-2-x(x+3))=0$

$(x+2)(-2-x^2-3x)=0$

$(x+2)(-(x^2+3x+2))=0$

Now we will factor by splitting the middle term.

$-(x+2)(x^2+2x+x+2)=0$

$-(x+2)(x+2)(x+1)=0$

$-(x+2)^2(x+1)=0$

Now we will use zero product property and solve for x as:

$-(x+2)^2=0,(x+1)=0$

$(x+2)=0,(x+1)=0$

$x=-2,x=-1$

Now, we will check for extraneous solutions.

We can see that $x=-2$ will make both denominators zero because our function is not defined at $x=-2$ .

Therefore, the solution for our given equation is $x=-1$ .

7 0

4 years ago

Read 2 more answers