At what point does y = 3x -2 intercept :

I do not know the answer for this, what is the answer?

guapka [62]

Answer:

B. translation, then reflection.

Step-by-step explanation:

translation means moving the shape without flipping or rotating it.

reflection means flipping it like in a mirror

5 0

2 years ago

Help please. this doesn't make sense

saveliy_v [14]

Answer:

a = 39.2650873379cm

b = 11.5529459767cm

Step-by-step explanation:

a - Use the area of a triangle equation - 1/2 * a * b * Sin(C)

1/2 * 8 * 10 * Sin(79) = 39.2650873379cm

b - Use the cosine rule - c^2 = a^2 + b^2 - (2 * a * b * Sin(C)

64 + 100 - (2 * 8 * 10 * Cos(79)) = 133.4705607398

c = square root of 6.9396506484 = 11.5529459767cm

Put to however many significant figures / decimal places required.

:)

5 0

2 years ago

When a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constan

vitfil [10]

Answer:

$a = 6m/s^2$

Step-by-step explanation:

Given

When mass = 4kg; Acceleration = 15m/s²

Required

Determine the acceleration when mass = 10kg, provided force is constant;

Represent mass with m and acceleration with a

The question says there's an inverse variation between acceleration and mass; This is represented as thus;

$a\ \alpha\ \frac{1}{m}$

Convert variation to equality

$a = \frac{F}{m}$ ; Where F is the constant of variation (Force)

Make F the subject of formula;

$F = ma$

When mass = 4kg; Acceleration = 15m/s²

$F = 4 * 15$

$F = 60N$

When mass = 10kg; Substitute 60 for Force

$F = ma$

$60 = 10 * a$

$60 = 10a$

Divide both sides by 10

$\frac{60}{10} = \frac{10a}{10}$

$a = 6m/s^2$

<em>Hence, the acceleration is </em> $a = 6m/s^2$ <em />

4 0

3 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

2 years ago

#1 Identify the characteristics of the quadratic graph

shusha [124]

Vertex: (1,-3)
Y intercept: -3
Axis of symmetry x=1
Domain all real numbers
I only got these the rest I don’t know because I just started learning it but these are correct.

8 0

3 years ago