Question 5 (1 point)

Mathematics

1 answer:

Leokris [45]2 years ago

8 0

A: Add up to 180 degrees

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Look at the picture<br><br>

Fantom [35]

Answer:

The x-intercept is 2.

Step-by-step explanation:

$2 \sqrt[3]{x - 10} + 4 = 0$

$2 \sqrt[3]{x - 10} = - 4$

$\sqrt[3]{x - 10 } = - 2$

$x - 10 = - 8$

$x = 2$

3 0

1 year ago

Solve for y -5y+8=-3y+10 what is the answer

masha68 [24]

-5y + 8 = -3y + 10
We need to get y on one side.
Add 3y to both sides.
-2y + 8 = 10
We need to get 2y by itself.
Subtract 8 from both sides.
-2y = 2
Divide both sides by -2 to solve for y.
y = -1

I hope this helps!

8 0

2 years ago

Read 2 more answers

Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8

Dima020 [189]

Let $f(x,y,z)=x+y+z-8e^{xyz}$ . The tangent plane to the surface at (0, 0, 8) is

$\nabla f(0,0,8)\cdot(x,y,z-8)=0$

The gradient is

$\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)$

so the tangent plane's equation is

$(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8$

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by $t$ , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

$(1,1,1)t+(0,0,8)=(t,t,t+8)$

or $x(t)=t$ , $y(t)=t$ , and $z(t)=t+8$ .

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)