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3 years ago

How do u solve 4(y-3)+9-3y

2 answers:

5 0

If you need to find the y, and the
4(y-3)+9-3y=0, then
we use PEMDAS, parenthesis first
4y-12+9-3y=0
4y-3y-12+9=0
y-3=0 add 3 on both sides
y=3

Komok [63]3 years ago

3 0

$4(y-3)+9-3y=0 \\ 4y-12+9-3y=0 \\ y-3=0 \\ y=3$

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Help! How would I solve this trig identity?

NeTakaya

Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

$sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}$

Then the identity can be rewritten as:

$sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\$

Now we can multiply both sides by cos⁴(x) to get:

$\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)$

Now we can use the identity:

sin²(x) + cos²(x) = 1

$1 - cos^2(x) = sin^2(x)*(sin^2(x) + cos^2(x)) = sin^2(x)\\\\1 = sin^2(x) + cos^2(x) = 1$

Thus, the identity was proven.

If you want to learn more about trigonometric identities:

brainly.com/question/7331447

#SPJ1

7 0

1 year ago

Plz help this is due today NO LINKS OR GROSS PICTURES OR I Will REPORT And please show work

aniked [119]

Answer:

2280 the volume

Step-by-step explanation:

So one box = 480

So then what about 6 boxes?

you do 480 x 6 = 2880

480+ 480+ 480+ 480+480+480= 2880

8 0

2 years ago

I need help!!!!!<br> #1<br> 5 1/2 + 0.6x=16<br> #2<br> 5/9 divided by x = 3 3/4<br> Thanks!!!!!!!1

nikdorinn [45]

Use the internet ok!!!!!!

4 0

3 years ago

Read 2 more answers

If one supplementary angle measure 51 degrees the other angle measures 39 degrees?

mojhsa [17]

The answer to your question is false

7 0

3 years ago

(x+5)(√ 1-x)<br> what is the product of this composite function?

Radda [10]

Answer:

$x\sqrt{1-x}+5\sqrt{1-x}$

Step-by-step explanation:

$(x+5)(\sqrt{1-x} )\\\\=\left(\sqrt{1-x}\right)\left(x+5\right)\\\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\\\a=\left(\sqrt{1-x}\right),\:b=x,\:c=5\\\\=\left(\sqrt{1-x}\right)x+\left(\sqrt{1-x}\right)\times\:5\\\\=x\left(\sqrt{1-x}\right)+5\left(\sqrt{1-x}\right)\\\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\\\=x\sqrt{1-x}+5\sqrt{1-x}$

6 0

3 years ago