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3 years ago

3. What’s the answer to this question asap!

Mathematics

1 answer:

icang [17]3 years ago

3 0

Because you are required to get a positive number of centimetres we use positive value of x.

$\mid6.75-x\mid$

The answer is B: The length of a part must be greater than 6.5 cm.

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With this diagram, what could be the values of q and r ?

neonofarm [45]

Q and r equals 13 because 13 and q match and r and 13 Mach in the high quality of mathematics

4 0

3 years ago

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Solve. y=2x-6 3x+y=4 x= y=

GenaCL600 [577]

Answer:

QUESTION 1

The given system of equation,

We simplify to get.

The correct answer is option A.

A.) (4, −2)

x = -7

y = 2

x = 5

y = 14

QUESTION 2

The given system of equations is

and

We make y the subject in equation (2) to get,

We put equation (3) into equation (1) to obtain,

We group like terms to get,

This implies that,

we divide through by to get,

Hence the x-coordinate is

QUESTION 3

The given system is

and

We make the subject in equation (2)

We put equation (3) into equation (1) to obtain,

We expand the bracket to get,

Group like terms to get,

We simplify to get;

This implies that,

Therefore the y-coordinate is 14.

Step-by-step explanation:

7 0

2 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5C%3A%20If%20%20%5C%3A%20y%20%3D%20%20%7Be%7D%5E%7Bx%7D%20%20-%20%20%5Cfrac%7B1%7D

VLD [36.1K]

Answer:

Given That: 

${ \large \longrightarrow{ \rm{ y = {e}^{x} - \frac{1}{x} + { log_{e}x }}}}$

On differentiating partially w.r.t. x, we get: 

${\large { \longrightarrow{ \rm{\frac{d}{dx} y = \frac{d}{dx} \left( {e}^{x} - \frac{1}{x} + log_ex \right)}}}}$

${ \large{ \longrightarrow { \rm{\frac{dy}{dx} = \frac{d}{dx} {e}^{x} - \frac{d}{dx} \frac{1}{x} + \frac{d}{dx} log_ex}}}}$

${ \large{ \longrightarrow { \rm{ \frac{dy}{dx} = {e}^{x} - \frac{d}{dx} {x}^{ - 1} + \frac{1}{x} }}}}$

${ \large{ \longrightarrow{ \rm{ \frac{dy}{dx} = {e}^{x} - [ - {x}^{ - 1 - 1} ] + \frac{1}{x} }}}}$

${ \large{ \longrightarrow{ \rm{ \frac{dy}{dx} = {e}^{x} - [ - {x}^{ - 2} ] + \frac{1}{x} }}}}$

Hence: 

${ \large \longrightarrow{ \green{ \boxed{{ \rm{ \frac{dy}{dx} = {e}^{x} + \frac{1}{ {x}^{2} } + \frac{1}{x} }}}}}}$

$\:$

Learn More: 

${\pink{\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}}}$

3 0

2 years ago

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- 4x+ 3y+x-2z+2y simplified

Klio2033 [76]

Answer:

-3x-2x+5y

hope this helps!

7 0

2 years ago

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