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3 years ago

I'LL GIVE YOU 35 POINTSSS HELP MEEE I NEED ASSISTANCE

Mathematics

1 answer:

Furkat [3]3 years ago

3 0

I hope this helps you

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Y > 4x + 3<br><br> y < -2x - 9

Zielflug [23.3K]

Answer:

x > 2

Step-by-step explanation:

y < 4x +3

y > -2x - 9

-2x -9 < 4x +3

-6x < -12

IN CASES WHERE YOU DIVIDE A NEGATIVE NUMBER,DON'T FORGET TO CHANGE THE SIGN !!!

x > 2

4 0

3 years ago

Need help, my teacher is not great at explaining.

satela [25.4K]

I cant really see the words

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3 years ago

X – 4x < 5x + 16 solve for x

Margaret [11]

Answer:

x>-2

Step-by-step explanation:

x-4x<5x+16

Subtract 5x from both sides:

x-4x-5x<16

Combine like terms:

-8x<16

Divide both sides by 8:

-x<2

Divide both sides by -1:

x>-2

Hope this helps!

3 0

3 years ago

Given sin(−θ)=−1/6 and tanθ=−√35/35 What is the value of cosθ?

murzikaleks [220]

Answer:

$cos(\theta)=-\frac{\sqrt{35} }{6}$

Step-by-step explanation:

Recall the negative angle identity for the sine function:

$sin(- \theta)=-sin(\theta)$

Then, we can find the value of $sin(\theta)$ :

$sin(\theta)=-sin(-\theta)\\sin(\theta) =-(-\frac{1}{6} )\\sin(\theta)= \frac{1}{6}$

Now recall the definition of the tangent function:

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

Therefore, now that we know the value of $sin(\theta)$ , we can solve in this equation for $cos(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\-\frac{\sqrt{35} }{35} =\frac{1/6}{cos(\theta)} \\cos(\theta)=-\frac{\frac{1}{6} }{\frac{\sqrt{35} }{35} } \\cos(\theta)=-\frac{35}{6\,\sqrt{35} } \\cos(\theta)=-\frac{\sqrt{35} }{6}$

5 0

2 years ago

Yara has 2 rectangular gardens in her backyard. The smaller garden is 1/2 foot long and 2/3

dimulka [17.4K]

Dimension of smaller garden :

l = 1/2 ft.

b = 2/3 ft.

Dimension of bigger garden :

L = 4 ft.

Let , breadth be x ft.

We know , area is given by :

Area = L×B.

Area of small garden = $\dfrac{1}{2}\times\dfrac{2}{3}=\dfrac{1}{3}\ ft^2$

Area of big garden $=4\times x=4x\ ft^2$

Hence, this is the required solution.

5 0

2 years ago