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3 years ago

Solve for x x-2=4+9z

Mathematics

2 answers:

weeeeeb [17]3 years ago

4 0

The answer will be X=6+9z

hodyreva [135]3 years ago

3 0

Answer:

x=6+9z

Step-by-step explanation:

x-2=4+9z

x-2+2=4+2+9z

x=6+9z

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How do I turn this equation into standard form? 8x=-5y+3

NARA [144]

Answer:

Step-by-step explanation:

To write a linear equation in standard form, move each variable term to the left side of the equation and simplify.

Ax+By=C

Since x is on the right side of the equation, switch the sides so it is on the left side of the equation.

8x−3=y

Move y to the left side of the equation because it contains a variable.

8x−3−y=0

Move 3 to the right side of the equation because it does not contain a variable.

8x−y=3

8 0

3 years ago

CAN SOMEONE PLEASE HELP ME WITH MY MATH ASAP PLEASE!!!!

Ivanshal [37]

Answer:

Step-by-step explanation:

Hello,

The two triangles are similar and x is positive, different from 0, so we can write.

$\dfrac{x}{9}=\dfrac{12}{x}\\\\x^2=12\times 9\\\\ x = \sqrt{4\times 3 \times 9}=2\times 3 \times \sqrt{3}\\\\ \boxed{\sf \bf x=6\sqrt{3} }$

Thanks

8 0

3 years ago

PLEASE HELP ME SEE IF MY ANSWERS ARE CORRECT!!!!!!!

Lena [83]

So first one
'how many solutions does 2x-y=-5 and 2x+y=5 have?'
add and get
2x-y-5
plus
2x+y=5

equals
2x+2x+y-y=5-5
4x=0
x=0 always
solve for y
4(0)+y=5
y=5
the solution is (0,5)
only ONE solution

one way is to subsitute
just remember that it is in (x,y) form so
the pont (1,2) means that 1 solution is x=1 and y=2 so subsitute and find that
the first one is the answer you are correct

just look at the graph
the solution is the intersection
it seems to be at a point that is 3 units to the right and -6 units up (6 units down)
so the solution is (3,-6)
yo are corect

subsitution
y=y
therefor
the answe ris (-4,-14) if you did the math correctly

#8 is correct

# 9 is correct

# 10 the answe ris bananas=0.40 pears=0.60

the last one you got it wrong, remember to check your answer to the graph for commonsense
then answer is (-2,5) and (1,2)

8 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .

$\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2$ .

$\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2$ .

Solve this equation for $u$ :

$\displaystyle \frac{u^{2} + 1}{u^{2}} = 2$ .

$u^{2} + 1 = 2 u^{2}$ .

$u^{2} = 1$ .

Given that $0 \le u \le 1$ , $u = 1$ is the only possible solution.

$\cos^{2}{x} = 1$ ,

$x = k \pi$ , where $k\in \mathbb{Z}$ (i.e., $k$ is an integer.)

Given that $0 \le x < 2\pi$ ,

$0 \le k$ .

$k = 0$ or $k = 1$ . Accordingly,

$x = 0$ or $x = \pi$ .

8 0

2 years ago

Read 2 more answers

The Brown family needs to rent a truck for their upcoming move. Express Movers will charge them $25 for the first day and $0.80

ankoles [38]

Express Movers equation would be y= 25 + .8x
Smith equation would be y= 35 + .6x
if you went 14 miles with Movers, you would go 1 mile with Smith for the same price.

8 0

2 years ago