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3 years ago

Which of these numbers is prime? 25, 26, 35, 39, 43

Mathematics

2 answers:

DENIUS [597]3 years ago

6 0

The number the is a primer is 43

ELEN [110]3 years ago

4 0

Answer:

The only number prime is 43

Step-by-step explanation:

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what is the slope intercept form of a line that is perpendicular to y = x + 3 and passes through point (2, -4)

Assoli18 [71]

Answer:

The equation of the perpendicular line is y = -x - 2

Step-by-step explanation:

* Lets revise the form of the slope intercept for

- The slope intercept form is y = mx + b, where m is the slope of

the line and b is the y-intercept

* Now lets revise the relation between the slopes of the

perpendicular lines

- If two lines are perpendicular, then the product of their slopes is -1

# Ex: If line L has slope m1 and line K has slope m2, and L ⊥ K

∴ m1 × m2 = -1

∴ m2 = -1/m1

* Now lets solve the problem

- We need to find the equation of the line which is perpendicular to

the line whose equation is y = x + 3 and passes through point (2 , -4)

- Find the slope of the given equation

∵ y = x + 3

- In this form the slope is the coefficient of x

∴ m = 1

- Find the slope of the perpendicular line

∵ The slope of the perpendicular line = -1/m

∴ The slope of it = -1/1 = -1

- Write the equation of the line with the value of the slope

∴ y = -x + b

- To find the value of b substitute x , y in the equation by the x and

y of the given point

∵ The line passes through point (2 , -4)

∵ y = -x + b

∴ -4 = -1(2) + b

∴ -4 = -2 + b ⇒ add 2 for both sides

∴ b = -2

- Write the equation with the value of b

∴ y = -x - 2

3 0

3 years ago

78 is 15% of what number? Solve using an equation. Show your work.

Phoenix [80]

Answer:

520

Step-by-step explanation:

78 = 15% × (what number)

78/0.15 = (what number) = 520

78 is 15% of 520.

4 0

3 years ago

Caroline knows the height and the required volume of a cone-shaped vase she’s designing. Which formula can she use to determine

timama [110]

Answer:

Option B. $r=\sqrt{\frac{3V}{\pi h}}$

Step-by-step explanation:

we know that

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2} h$

Solve for the radius r

That means-----> isolate the variable r

Multiply by 3 both sides

$3V=\pi r^{2} h$

Divide by $(\pi h)$ both sides

$r^{2}=\frac{3V}{\pi h}$

square root both sides

$r=\sqrt{\frac{3V}{\pi h}}$

4 0

3 years ago

PLS HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

scoray [572]

Answer:

Clam down jamal don't pull out the 9 X_X

Step-by-step explanation:

3 0

3 years ago

HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im

GrogVix [38]

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for $f:\mathbb R^2\to\mathbb R$ , we have the partial derivative with respect to $x$ defined as

$\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h$

The derivative at (0, 0) is then

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h$

• By definition of $f$ , $f(0,0)=0$ , so

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}$

• Expanding the tangent in terms of sine and cosine gives

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}$

• Introduce a factor of $g(h,0)$ in the numerator, then distribute the limit over the resulting product as

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h$

• The first limit is 1; recall that for $a\neq0$ , we have

$\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1$

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h$

and this is exactly the partial derivative of $g$ with respect to $x$ .

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)$

For the same reasons shown above,

$\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)$

(b) To show that $f$ is differentiable at (0, 0), we first need to show that $f$ is continuous.

• By definition of continuity, we need to show that

$\left|f(x,y)-f(0,0)\right|$

is very small, and that as we move the point $(x,y)$ closer to the origin, $f(x,y)$ converges to $f(0,0)$ .

We have

$\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}$

The first expression in the product is bounded above by 1, since $|\sin(x)|\le|x|$ for all $x$ . Then as $(x,y)$ approaches the origin,

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0$

So, $f$ is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which $f(x,y)$ changes as we move the point $(x,y)$ closer to the origin, given by

$\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,$

approaches 0.

Just like before,

$\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}$

and this converges to $g(0,0)=0$ , since differentiability of $g$ means

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0$

So, $f$ is differentiable at (0, 0).

3 0

3 years ago