All categories

2 years ago

The product of a real number and another real number that is 6 less than the first equals 6. What are the numbers?

Mathematics

2 answers:

malfutka [58]2 years ago

5 0

1? Or dividend or multiplication

UNO [17]2 years ago

5 0

Answer:

$3+\sqrt{15} \\3-\sqrt{15}\\$

Step-by-step explanation:

if ax=6 and a = x-6...

x(x-6) = 6

Completing the square is a process such that $(x^2+2bx)=(x+b)^2-b^2$

You might be interested in

Select all expressions that represent the area of the given triangle

Mice21 [21]

Answer:

A=1/2(12x16)

Step-by-step explanation:

because it's the area of the square divided by two

3 0

3 years ago

When creating lines of best fit, do you believe that estimation by inspection of the equation is best or do you think it should

Alex73 [517]

Answer:

Estimation by inspection is better than trying to determine the line of best fit exactly

i) For a scatter plot : The use of estimation by inspection

ii) For a straight line graph : The exact determination method

Step-by-step explanation:

To create lines of best fit the estimation by inspection is better than trying to determine the line of best fit exactly .

This is because line of best fit only shows the trend of the data and in most cases it doesn't have to start from origin.

Scenarios :

i) For a scatter plot : The use of estimation by inspection

ii) For a straight line graph : The exact determination method

8 0

3 years ago

The area of a rectangle court is 433.37 Square meters and the length of the court is 28.7 meters . What is width of the court?

Eduardwww [97]

Answer:

w=15.1m

Step-by-step explanation:

l=28.7

A=433.37

7 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

Suppose you have an equal sided spinner that is numbered 1 thru 5 and another spinner with 4 equal sections with the colors red,

DanielleElmas [232]

Answer:6

778

Step-by-step explanation:

hh8yyyyyyyyyyyyyy

8 0

3 years ago

Read 2 more answers