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3 years ago

0.4(2x + 0.3) = 1/3 (6x - 7.2)

Mathematics

2 answers:

GenaCL600 [577]3 years ago

5 0

What are u looking for ?

Hitman42 [59]3 years ago

4 0

Answer:

x = 20.1

Step-by-step explanation:

First up you need to multipy the numbers on the outside with the ones on the inside

This would mean that

0.4(2x + 0.3) = 1/3(6x - 7.2) will change to

0.8x + 0.12 = 2x - 2.4

You just have to solve for x now:

2.4 + 0.12 = 2x - 0.8x

2.52 = 1.2x

x = 2.1

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If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?

eduard

Answer:

$cos(\alpha+\beta)=-\frac{84}{85}$

Step-by-step explanation:

we know that

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

Remember the identity

$cos^{2} (x)+sin^2(x)=1$

step 1

Find the value of $sin(\alpha)$

we have that

The angle alpha lie on the III Quadrant

The values of sine and cosine are negative

$cos(\alpha)=-\frac{8}{17}$

Find the value of sine

$cos^{2} (\alpha)+sin^2(\alpha)=1$

substitute

$(-\frac{8}{17})^{2}+sin^2(\alpha)=1$

$sin^2(\alpha)=1-\frac{64}{289}$

$sin^2(\alpha)=\frac{225}{289}$

$sin(\alpha)=-\frac{15}{17}$

step 2

Find the value of $cos(\beta)$

we have that

The angle beta lie on the IV Quadrant

The value of the cosine is positive and the value of the sine is negative

$sin(\beta)=-\frac{4}{5}$

Find the value of cosine

$cos^{2} (\beta)+sin^2(\beta)=1$

substitute

$(-\frac{4}{5})^{2}+cos^2(\beta)=1$

$cos^2(\beta)=1-\frac{16}{25}$

$cos^2(\beta)=\frac{9}{25}$

$cos(\beta)=\frac{3}{5}$

step 3

Find cos⁡ (α + β)

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

we have

$cos(\alpha)=-\frac{8}{17}$

$sin(\alpha)=-\frac{15}{17}$

$sin(\beta)=-\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute

$cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})$

$cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}$

$cos(\alpha+\beta)=-\frac{84}{85}$

4 0

3 years ago

18. m21 S 21. Guided

xz_007 [3.2K]

Answer:

Step-by-step explanation:

say what is this about the phone

3 0

2 years ago

(PLEASE ANSWER 50 POINTS)Write an equation of a line slope-intercept form with the given slope and y-intercept. Slope:3, y-inter

Levart [38]

Y=3x+2

the slope (mx) is 3
the y-intercept (b) is 2.

the equation form is y=mx+b

8 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

marissa [1.9K]

Answer:

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

Step-by-step explanation:

If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -

x = ( 30 cos 20° )( time ),

y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2

To determine " ( 30 cos 20° )( time ) " you would do the following calculations -

( x = 30 * 0.93... = ( About ) 28.01t

This represents our horizontal distance, respectively the vertical distance should be the following -

y = 30 * 0.34 - 4.9t^2,

( y = ( About ) 10.26t - 4.9t^2 + 2

In other words, our solution should be,

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

These are are parametric equations

5 0

3 years ago

What is the solution set for -4x - 10 ≤ 2?

hichkok12 [17]

Answer:

x > -3

Step-by-step explanation:

- 4x - 10 < 2

+ 10 < + 10

- 4x < 12

-4 -4

x > -3

I think the sign flips because the negative sign is w/the X.... If it doesn't, sorry 4 being wrong...

8 0

3 years ago