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2 years ago

Write an equation in slope-intercept form for this line.

Mathematics

1 answer:

sineoko [7]2 years ago

7 0

Answer:

y = $\frac{3}{2}$ x - 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = $\frac{3}{2}$ , then

y = $\frac{3}{2}$ x + c ← is the partial equation

To find c substitute (- 4, - 7 ) into the partial equation

- 7 = - 6 + c ⇒ c = - 7 + 6 = - 1

y = $\frac{3}{2}$ x - 1 ← equation of line

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Find sin(a+b) if tan(a)=7/24 where a is in the third quadrant

ludmilkaskok [199]

The complete question in the attached figure

we have that
tan a=7/24 a----> III quadrant
cos b=-12/13 b----> II quadrant
sin (a+b)=?

we know that
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

step 1
find sin b
sin²b+cos²b=1------> sin²b=1-cos²b----> 1-(144/169)---> 25/169
sin b=5/13------> is positive because b belong to the II quadrant

step 2
Find sin a and cos a
tan a=7/24
tan a=sin a /cos a-------> sin a=tan a*cos a-----> sin a=(7/24)*cos a
sin a=(7/24)*cos a------> sin²a=(49/576)*cos²a-----> equation 1
sin²a=1-cos²a------> equation 2
equals 1 and 2
(49/576)*cos²a=1-cos²a---> cos²a*[1+(49/576)]=1----> cos²a*[625/576]=1
cos²a=576/625------> cos a=-24/25----> is negative because a belong to III quadrant
cos a=-24/25
sin²a=1-cos²a-----> 1-(576/625)----> sin²a=49/625
sin a=-7/25-----> is negative because a belong to III quadrant

step 3
find sin (a+b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin a=-7/25
cos a=-24/25
sin b=5/13
cos b=-12/13
so
sin (a+b)=[-7/25]*[-12/13]+[-24/25]*[5/13]----> [84/325]+[-120/325]
sin (a+b)=-36/325

the answer is
sin (a+b)=-36/325

8 0

2 years ago

Y=5x-4

anyanavicka [17]

Answer:

$\displaystyle -5x + y = -92\:or\:y = 5x - 92$

Step-by-step explanation:

−12 = 5[16] + b

$\displaystyle -92 = b \\ \\ y = 5x - 92$

If you want it in Standard Form:

y = 5x - 92

- 5x - 5x

__________

$\displaystyle -5x + y = -92$

I am joyous to assist you anytime.

7 0

2 years ago

polynomial function of degree 4 with -1 as a zero of multiplicity 3 and 0 as a zero of multiplicity 1

iren [92.7K]

$\bf \begin{cases}
x=-1\implies &x+1=0\\
x=0\implies &x=0
\end{cases}
\\\\\\
\stackrel{\textit{multiplicity of 3}}{(x+1)^3}~~\stackrel{\textit{multiplicity of 1}}{(x)^1}=\stackrel{original~polynomial}{0}
\\\\\\
(x^3+3x^2+3x+1)(x)=\stackrel{y}{0}\implies x^4+3x^3+3x^2+x=y$

6 0

2 years ago

Which transformations can be used to map a triangle with vertices A(2, 2), B(4, 1), C(4, 5) to A’(–2, –2), B’(–1, –4), C’(–5, –4

jek_recluse [69]

Notice that every pair of point (x, y) in the original picture, has become (-y, -x) in the transformed figure.

Let ABC be first transformed onto A"B"C" by a 90° clockwise rotation.

Notice that B(4, 1) is mapped onto B''(1, -4). So the rule mapping ABC to A"B"C" is (x, y)→(y, -x)

so we are very close to (-y, -x).

The transformation that maps (y, -x) to (-y, -x) is a reflection with respect to the y-axis. Notice that the 2. coordinate is same, but the first coordinates are opposite.

ANSWER:

"a 90 clockwise rotation about the origin and a reflection over the y-axis"

5 0

2 years ago

Read 2 more answers

I am doing this test and I need questions questions ASAP I will give brainiest to whoever answers it correctly

Troyanec [42]

Answer:

The correct answer there is the first option

6 0

2 years ago