Sixty percent of the days in the desert were hot and dry . Out of the 365 days, how many were hot and dry?

Fill in the blank to make the expression a perfect square:

tigry1 [53]

Answer:

64

Step-by-step explanation:

So here you're just completing the square. the equation you gave is simply: $n^2+16n+c$ where c is the unknown value we're solving for. Whenever you complete the square, you add (b/2)^2

The reason for this, is because whenever you write a binomial as a perfect square it's in the form: $(x+b)^2$ and this binomial expands out to become: $x^2+2bx+b^2$

If we write the second term of the binomial as b/2 we get:

$(x+\frac{b}{2})^2=x^2+2(\frac{b}{2})x+(\frac{b}{2})^2$

which simplifies to:

$(x+\frac{b}{2})^2=x^2+bx+(\frac{b}{2})^2$

and as you can see the last term is (b/2)^2, which is why we need to add that part for it to be a perfect square.

So we would need to add (16/2)^2 = 8^2 = 64

This way, we can express it as a perfect square binomial: $(n+8)^2$ which expands out to: $n^2+2(8)(n)+8^2 = n^2+16n+64$

7 0

2 years ago

Find the pattern and identify the next 3 terms in the sequence 5, -10, 20...

Marta_Voda [28]

Answer:

-40, 80, -160

Step-by-step explanation:

5(-2)^n-1

where n is the position

6 0

3 years ago

If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction

valentina_108 [34]

Answer:

a) $\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$ .

b) $Du_{f}(2,4,0) = -\frac{8}{\sqrt{11}}$

Step-by-step explanation:

Given a function $f(x,y,z)$ , this function has the following gradient:

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

(a) find the gradient of f

We have that $f(x,y,z) = x\sin{yz}$ . So

$f_{x}(x,y,z) = \sin{yz}$

$f_{y}(x,y,z) = xz\cos{yz}$

$f_{z}(x,y,z) = xy \cos{yz}$ .

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

$\nabla f(2,4,0) = \sin{0}\mathbf{i} + 0\cos{0}\mathbf{j} + 8 \cos{0}\mathbf{k}$ .

$\nabla f(2,4,0) = 0i+0j+8k=(0,0,8)$

The vector is $v = i + 3j - k = (1,3,-1)$

To use v as an unitary vector, we divide each component of v by the norm of v.

$|v| = \sqrt{1^{2} + 3^{2} + (-1)^{2}} = \sqrt{11}$

So

$v_{u} = (\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}})$

Now, we can calculate the scalar product that is the directional derivative.

$Du_{f}(2,4,0) = (0,0,8).(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}) = -\frac{8}{\sqrt{11}}$

6 0

3 years ago

Suppose a friend is having difficulty solving -2(q-5) > -3(q+1). Explain how to solve the inequality, showing all the necessa

IgorC [24]

Firstly, use the distributive property of multiplication (A(B + C) = A×B + A×C) on -2(q - 5) and -3(q + 1): $-2q+10>-3q-3$

Next, apply the addition property of equality (whatever you add to one side you have to add the same quantity to the other), and add 3q on both sides: $q+10>-3$

Lastly, apply the subtraction property of equality (whatever you subtract on one side you have to subtract the same amount on the other side), and subtract 10 on both sides. Your final answer will be $q>-13$ 

8 0

3 years ago

I need this done asap will give brainliest

lapo4ka [179]

2,4 and I am not too sure but I think -4

3 0

3 years ago