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2 years ago

Sinx + cosx= 1 then find the value of x

1 answer:

4 0

$1√2sinx+1√2cosx=1√2.

This can be written as

cos(x−π4)=cos(π4)

The general solution of this equation ls

x−.π4=2nπ±π4,n=0,±1,±2,...,

So, x=2nπandx=(4n+1)π2,n=0,±1,±2,±3....

 ANSWER: x=2nπandx=(4n+1)π2,n=0,±1,±2,±3....$

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jennifer walks 8 miles in the same time she can jog 13 miles. if her jogging speed is 8mph faster than her walking speed, how fa

Elanso [62]

Answer:

12.8mph

Step-by-step explanation:

Let 'w' represent her walking speed

->if her jogging speed is 8mph faster than her walking speed

Jogging speed= w+8

Using the time equation i.e time = dist/speed

For walking when distance is 8miles

time = dist / speed= 8/w hrs

For jogging when distance is 13miles

time = dist / speed= 13/(w+8) hrs

Equating the time

jog time = walk time

13/(w+8) = 8/w

By cross multiplying it

13w = 8w + 64

5w = 64

w=64/5

w=12.8mph

Therefore, 12.8mph is her walking speed.

5 0

2 years ago

Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

Here we want to find all the points such that

$\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)$

where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

8 0

2 years ago

Solve for x: x5 + x4 − 7x3 − 7x2 − 144x − 144 = 0

motikmotik

$x^5 + x^4 - 7x^3 - 7x^2 -144x - 144 = 0 \\x^4(x+1)-7x^2(x+1)-144(x+1)=0\\(x^4-7x^2-144)(x+1)=0\\\\x+1=0\Rightarrow x=-1\\\\x^4-7x^2-144=0\\x^4-16x^2+9x^2-144=0\\x^2(x^2-16)+9(x^2-16)=0\\(x^2+9)(x^2-16)=0\\(x^2+9)(x-4)(x+4)=0\\\\x^2+9=0\vee x-4=0 \vee x+4=0\\x=4 \vee x=-4\\\\x\in\{-4,-1,4\}$

6 0

3 years ago

Read 2 more answers

A freight train is half a mile long. It approaches a tunnel traveling 60mi/hr. If the tunnel is 1 mile long, how long will it ta

MrRa [10]

Answer:60

Step-by-step explanation:I seen the answer on bill bye the math guy

6 0

2 years ago

Read 2 more answers

Pleaseeee answer correctly <br> Will mark Brianliest !!! HELP !

Paha777 [63]

Answer: 11

The code would print 11 as x=x-4 basically sets x to what x was and subtracts it by 4 which means 15 was its previous value and it takes 4 away from it so it is now 11. Then because the loop condition is x=11 the loop condition has been met and will now go and display x.

8 0

3 years ago