UV=TW. Find VX will give you brainliest if it’s correct

Write each equation in slope-intercept form of the equation of a line. Underline the slope and circle the y-intercept in each eq

VARVARA [1.3K]

$\bf 3x + 4y - 12 = 0\implies 4y-12 = -3x\implies 4y=-3x+12 \\\\\\ y = \cfrac{-3x+12}{4}\implies y = \cfrac{-3x}{4}+\cfrac{12}{4} \\\\\\ y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{4}}x\stackrel{\stackrel{b}{\downarrow }}{+3} \qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

5 0

3 years ago

Read 2 more answers

MATH HELP PLZ!!!

RoseWind [281]

Answer:

a) $tan (157.5) = \frac{1-cos 315}{sin315}$

b)

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c)

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

d)

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

a)

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

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

b)

cos2∝ = 1- 2 sin² ∝

cos∝ = 1- 2 sin² ∝/2

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

Step(i):-

Given

$tan\alpha = \frac{sin\alpha }{cos\alpha }$

we know that trigonometry formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

Given

$tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }$

put ∝ = 315

$tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }$

multiply with ' 2 sin (∝/2) both numerator and denominator

$tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2} }{2sin(\frac{315}{2} cos(\frac{315}{2}) }$

Apply formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

now we get

$tan (157.5) = \frac{1-cos 315}{sin315}$

b)

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

put ∝ = 330° above formula

$sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}$

$sin^{2} (165) = \frac{1-cos (330) }{2}$

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c )

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

put ∝ = 315° above formula

$sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}$

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

d)

cos∝ = 1- 2 sin² ∝/2

put ∝ = 330°

$cos 330 = 1 - 2sin^{2} (\frac{330}{2} )$

cos 330° = 1- 2 sin² (165°)

3 0

3 years ago

543,000,000,000,000 in Scientific notation

Zielflug [23.3K]

Answer:

5.43×10^14 or

$5 .43 \times 10^{14}$

Step-by-step explanation:

Scientific notation, the number must always be less than 10, in this case 5.43. The exponent represents how much times I moved the decimal point to the left.

8 0

3 years ago

Which expression is equivalent to ^3 sqrt 1/1000 c^9 d^12

tamaranim1 [39]

Answer:

$\sqrt[3]{ \frac{1}{1000} {c}^{9} {d}^{12} } \\ = \frac{1}{10} {c}^{3} {d}^{4}$

5 0

3 years ago

The force F of attraction between two bodies varies jointly as the weights of the two bodies and inversely as the square of the

crimeas [40]

We have that F is proportional to m1. Hence, we have that F=m1*c when we change the first mass. We also have that F=m2*c when we only change the 2nd mass. Since F scales inversely with the square of the distance, we have that $F= \frac{c}{d^2}$ . Combining all of the facts above, we get that F=c* $\frac{m_1 m_2}{d^2}$ where c is a constant.

3 0

3 years ago