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pickupchik [31]
3 years ago
9

Find the number that belongs in the green box. Round your answer to the nearest tenth.

Mathematics
2 answers:
Andru [333]3 years ago
7 0

Answer:

1.2 (to one tenth)

Step-by-step explanation:

Hope that helps :-)

allsm [11]3 years ago
6 0

Answer:

Step-by-step explanation:

Remark

You are going to have to find the angle on the lower right before you can get the answer. Call the angle on the lower right = x

Equation

Sin(60)/7 = sin(x) / 3

Solution

3 * sin(60) / 7 = sin(x)

Sin(60) = 0.866

3*0.866 / 7 = sin(x)

sin(x) = 0.3712

x = sin-1(0.3712)

x = 21.79

Now calculate the 3rd angle call it y

y + 21.79 + 60 = 180

y + 81.79 = 180

y = 180 - 81.79

y = 98.21

Now use the sin law again.

Let ? = z

z/sin(98.21) = 7/sin(60)              Multiply both sides by sin(98.21)

z = 7 * sin(98.21) / sin(60)

z = 6.9282 / sin(60)

z = 8

? = 8

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Step-by-step explanation:

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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
3 years ago
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