All categories

2 years ago

Solve for x help thx

Mathematics

2 answers:

My name is Ann [436]2 years ago

7 0

Answer:

x = 7

Step-by-step explanation:

Given

3(2x - 9) = 15 ( divide both sides by 3 )

2x - 9 = 5 ( add 9 to both sides )

2x = 14 ( divide both sides by 2 )

x = 7

valentinak56 [21]2 years ago

5 0

X = 8
because 8 x 2 = 16
16 - 9 = 5
5 x 3 = 15

You might be interested in

Write an equation of the form y=a sin bx or y= a cos bx to describe the graph below.

kakasveta [241]

we conclude that the graphed equation is:

y = 4*cos(pi*x)

<h3></h3><h3>Which trigonometric equation is the one in the graph?</h3>

First, we can see that the graphed function is even, so we know that it wiill be a cosine.

We also can see that the maximum is 4 and the minimum is -4, so the amplitude is 4.

Then we have something like:

y = 4*cos(b*x)

To find the value of b, we can use the fact that the zeros of the function are at x = ±1/2

Then:

b*1/2 = pi/2

b = pi

Finally, we conclude that the graphed equation is:

y = 4*cos(pi*x)

If you want to learn more about trigonometric equations:

brainly.com/question/8120556

#SPJ1

4 0

11 months ago

Read 2 more answers

molle have 100098765 her mom gave her 800000 A week later she got 90000 Tow dAyS later she spent 900000 how much dose sHe HaVe

dangina [55]

Answer:

100088765

Step-by-step explanation:

7 0

2 years ago

What is 5- (t+3)= -1+2(t-3)

maks197457 [2]

Answer:

Step-by-step explanation:

t=3

7 0

1 year ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

<u>Prove that:</u>

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

<u>Proof: </u>

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

<u>Taking</u><u> </u><u>LHS</u>

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$