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3 years ago

How can I solve this equation, 6x + 3y = -12, as y=mx+b?

Mathematics

1 answer:

djyliett [7]3 years ago

5 0

Answer:

y=-2x-4

Step-by-step explanation:

6x + 3y=-12 subtract 6x to other side

3y=-12-6x divide 3 to every number

1y=-4-2x

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Mrs. Murphy bought balloons, garlands, paper lanterns, and banners to

kvasek [131]

Answer:

Banners: 4

Paper lanters: 3

Bags of Balloons: 6

Garlands: 11

Step-by-step explanation:

Banner: 4*5.25= $21

Paper lantern: 6.50*x=?

Balloons: 5*x=?

Garland: 3.50*x=?

Paper lantern: 145.50/6.50= $22.3

22.3/6.50= 3.4 round to 3

3 paper lanterns

Balloons: 145.50/5= $29.1

29.1/5= 5.8 round to 6

6 packs of balloons

Garland: 145.50/3.50= $41.5

41.5/10= 4.15

4.15+7=11.15 round to 11

11 garlands

4 0

3 years ago

In a game show, players play multiple rounds to score points. Each round has 3 times as many points available as the previous ro

Juli2301 [7.4K]

Answer:

$p = 25(3)^n$

Step-by-step explanation:

Given

Round 0 1 2 3

Points 25 75 225 675

Required

Determine the equation

Let n = rounds and p = points.

So, we have:

$(n_1,p_1) = (0,25)$

$(n_2,p_2) = (1,75)$

$(n_2,p_2) = (3,225)$

The above shows a geometric progression.

So, we solve using

$p = ab^n$

In $(n_1,p_1) = (0,25)$

n = 0 and p = 25

Substitute these values in $p = ab^n$

$25 = ab^0$

$25 = a*1$

$25 = a$

$a = 25$

In $(n_2,p_2) = (1,75)$

n = 1 and p = 75

Substitute these values in $p = ab^n$

$75 = ab^1$

$75 = ab$

Substitute 25 for a

$75 = 25*b$

Make b the subject

$b = \frac{75}{25}$

$b = 3$

So, we have:

$a = 25$ and $b = 3$

The equation $p = ab^n$ becomes

$p = 25(3)^n$

5 0

3 years ago

Molly is excited to have fresh basil at home. She buys a basil plant that is 90 millimeters tall and puts it on her kitchen wind

satela [25.4K]

Answer:

Molly's basil plant is now 17 centimeters

Step-by-step explanation:

90 millimeters=9 centimeters

20cm-12cm=8cm

9cm+8cm=17cm

hope this helped :)

5 0

2 years ago

Show that ∆ABC is a right triangle if A(0, 0), B(3, 2), and C(–2, 3). Find the slope m1 of AB and the slope m2 of AC and verify

Mariana [72]

First we'll show this is a right triangle by the Pythagorean Theorem:

$A(0,0), B(3,2), C(-2, 3)$

$AB^2=3^2+2^2=13$

$AC^2=(-2)^2+3^2=13$

$BC^2 = (-3 -3)^2 + (3 -2)^2 = 26$

Since $AB^2+AC^2=BC^2$ we have a right triangle, by the (converse to the) Pythagorean Theorem. We also see it's isosceles, AB=AC.

$\textrm{Slope of AB} = m_1 = \dfrac{2 - 0}{3 - 0} = \dfrac 2 3$

$\textrm{Slope of AC} = m_2 = \dfrac{3 - 0}{-2 - 0} = -\dfrac 3 2$

$m_1 m_2 = (\frac 2 3)(-\frac 3 2 ) = -1 \quad\checkmark$

That's the end of the homework but I'll go on a bit.

Another way to show perpendicularity, essentially the same as the other two, is by a zero as the dot product of the sides as vectors, differences between vertices.

Vector AB = B - A = (3,2)

Vector AC = C - A = (-2, 3)

AB · AC = 3(-2) + 2(3) = 0

A zero dot product means

AB ⊥ AC

8 0

3 years ago

Write an equation of the line that passes through (3, 1) and (0, 10).<br> y

const2013 [10]

Answer:

Step-by-step explanation:

(10 - 1)/(0 - 3) = -9/3 = -3

y - 1 = -3(x -3)

y - 1 = -3x + 9

y = -3x + 10

7 0

3 years ago