-x + 5y = -17 <br> -2x -5y = -4<br><br> using elimination

Alex is drawing rectangles with different areas on a centimetre grid he can draw 3 different rectangles with an area of 12cmsqua

Naily [24]

Answer:

Alex can only draw a rectangle with an area of 11 cm²

The reason for the answer is that 11 is a prime number

Step-by-step explanation:

We are told that Alex is drawing rectangles with different areas on a centimeter grid, and he can draw 3 different rectangles with an area of 12 cm²

We have the minimum grid length of centimeters = 1 cm, since we assume that only integers can be used.

That is to say:

12 = 1 x 12

12 = 2 x 6

12 = 3 x 4

These are the 3 different rectangles with an area of 12 cm²

Now we have to find how many rectangles Alex could draw with an area of 11 cm²

11 = 1 x 11

Therefore, only one factorization is possible. Which means that you can only draw a rectangle with an area of 11 cm²

The reason for the answer is that 11 is a prime number, which means only the common factor of 1 and the number itself, that is, 11.

7 0

3 years ago

A line segment has the endpoints (2 -5) and (-1 -1) what is its slope?

Ganezh [65]

I believe,
since... slope = (y2-y1)/(x2-x1)
so... slope = (-1+5)/(-1-2)
slope = (4)/(-3) or.. -.75

3 0

3 years ago

Round to the nearest hundredth 42.052

konstantin123 [22]

Answer:

To round 42.052 to the nearest hundredth consider the thousandths' value of 42.052, which is 2 and less than 5. Therefore, the hundredths' value of 42.052 remains 5.

5 0

3 years ago

Read 2 more answers

Circle―r = 4 cm find the area

rusak2 [61]

Are of circle = πr² = 22/7 *4*4 = 50.27cm²

6 0

3 years ago

Use De Moivre's theorem to write the complex number in trigonometric form: (cos(3pi/5) + i sin (3pi/5))^3

docker41 [41]

By De Moivre's theorem,

$\left(\cos\dfrac{3\pi}5+i\sin\dfrac{3\pi}5\right)^3=\boxed{\cos\dfrac{9\pi}5+i\sin\dfrac{9\pi}5}$

We can stop here ...

# # #

... but we can also express these trig ratios in terms of square roots. Let $x=\dfrac\pi5$ and let $c=\cos x$ . Then recall that

$\cos5x=c^5-10c^3\sin^2x+5c\sin^4x$

$\cos5x=c^5-10c^3(1-c^2)+5c(1-2c^2+c^4)$

$\cos5x=16c^5-20c^3+5c$

On the left, $5x=\pi$ so that

$16c^5-20c^3+5c+1=(c+1)(4c^2-2c-1)^2=0$

Since $c=\cos\dfrac\pi5\neq1$ , we're left with

$4c^2-2c-1=0\implies c=\dfrac{1+\sqrt5}4$

because we know to expect $\cos x>0$ . Then from the Pythagorean identity, and knowing to expect $\sin x>0$ , we get

$\sin x=\sqrt{1-c^2}=\sqrt{\dfrac58-\dfrac{\sqrt5}8}$

Both $\cos$ and $\sin$ are $2\pi$ -periodic, so that

$\cos\dfrac{9\pi}5=\cos\left(\dfrac{9\pi}5-2\pi\right)=\cos\left(-\dfrac\pi5\right)=\cos\dfrac\pi5$

and

$\sin\dfrac{9\pi}5=\sin\left(-\dfrac\pi5\right)=-\sin\dfrac\pi5$

so that the answer we left in trigonometric form above is equal to

$\cos\dfrac\pi5-i\sin\dfrac\pi5=\boxed{\dfrac{1+\sqrt5}4+i\sqrt{\dfrac58+\dfrac{\sqrt5}8}}$

7 0

3 years ago

-x + 5y = -17 -2x -5y = -4 using elimination