Mari used a thermometer to record temperatures of -3.4 Celsius and 1.6 Celsius . Which temperature in degrees is lower

1. Find the value of x A. 7 B. 4 C. 5 D. 6

Damm [24]

A I believe, sorry if I’m wrong

4 0

3 years ago

The equation of the tangent plane to the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 at the point (x0, y0, z0) can be written as xx0 a2

svetlana [45]

Answer:

The equation of tangent plane to the hyperboloid

$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=1$ .

Step-by-step explanation:

Given

The equation of ellipsoid

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

The equation of tangent plane at the point $\left(x_0,y_0,z_0\right)$

$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}=1$ ( Given)

The equation of hyperboloid

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$

F(x,y,z)= $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}[c^2}$

$F_x=\frac{2x}{a^2},F_y=\frac{2y}{b^2},F_z=-\frac{2z}{c^2}$

$(F_x,F_y,F_z)(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},-\frac{2z_0}{c^2}\right)$

The equation of tangent plane at point $\left(x_0,y_0,z_0\right)$

$\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)-\farc{2z_0}{c^2}(z-z_0)=0$

The equation of tangent plane to the hyperboloid

$\frac{2xx_0}{a^2}+\frac{2yy_0}{b^2}-\frac{2zz_0}{c^2}-2\left(\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}-\frac{z_0^2}{c^2}\right)=0$

The equation of tangent plane

$2\left(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}\right)=2$

Hence, the required equation of tangent plane to the hyperboloid

$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=0$

7 0

3 years ago

A ∠A and ∠ B ∠B are complementary angles. If m ∠ A = ( 2 x − 23 ) ∘ ∠A=(2x−23) ∘ and m ∠ B = ( x + 20 ) ∘ ∠B=(x+20) ∘ , then fin

son4ous [18]

Answer:

∠A = 99°

Step-by-step explanation:

Complementary angles sum to 180°, thus

∠A + ∠B = 180 ← substitute values

2x - 23 + x + 20 = 180

3x - 3 = 180 ( add 3 to both sides )

3x = 183 ( divide both sides by 3 )

x = 61

Hence

∠A = 2x - 23 = (2 × 61) - 23 = 122 - 23 = 99°

7 0

3 years ago

$250, 2.85%, 3 years Simple interest

rosijanka [135]

Ok so you put I = 250 x .0285 x 3 and it gives you 21,375 then you add 21,375 with 250 and it gives you 21,625 that world be your answer

8 0

3 years ago

Determine the equation of the line that passes through the given points. (If you have a graphing calculator, you can use the tab

expeople1 [14]

You can write the equation in point-slope form, which has the format y-ysubscript1=m(x-xsubscript1), with ysubscript1 and xsubscript1 being the y and x coordinates for a point on the line, and m being the slope.

Substitute a y and x coordinate into the equation so you have y-6=m(x-2)

Then find the slope so you can replace m. The slope formula is (ysubscript2-ysubscript1)/(xsubscript2-xsubscript1). Substitute the coordinates in so you have m=(16-6)/(4-2), which simplifies to 10/2 and then 5.

Now the equation is y-6=5(x-2)

If you want a different form, for example slope-intercept form, you can change it to that:

y-6=5(x-2)

y=5x-4

6 0

3 years ago