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Georgia [21]
3 years ago
9

Question 5 A B C D PLEASE HELP!!!!!!

Mathematics
1 answer:
Eddi Din [679]3 years ago
5 0

Answer:

a)x=-2.5 or x=1

Step-by-step explanation:

it's the answer

hope It helps

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4^2 + (10 - 2 × 3) ÷ 4<br><br><br> (i just need it to be explain thank you.)
tiny-mole [99]

Answer:

PEMDAS

Step-by-step explanation:

you can solve this, just remember pemdas

5 0
3 years ago
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How do i find the equation of the line?
Cerrena [4.2K]

Answer:

10x + 7y = 2

Step-by-step explanation:

Slope = (6 + 4)/(-4 - 3) = -10/7

Equation:

y + 4 = -10/7(x - 3)

7(y + 4) = - 10(x - 3)

7y + 28 = -10x + 30

10x + 7y = 30 - 28

10x + 7y = 2

6 0
4 years ago
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Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
4 years ago
Find the exact value of the following, giving your answers as a fraction. c) 2^-6
solmaris [256]
2^6=2*2*2*2*2*2=64

Since the exponent is negative, we put the 64 over 1

1/64
8 0
3 years ago
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What is y + 10 = -3(x + 10) in slope-intercept form?
Inga [223]

Answer:

y= -3x-40

Step-by-step explanation:

first distribute

y+10= -3x-30

than subtract 10

y= -3x-40

4 0
4 years ago
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