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3 years ago

Jace estimated the product of (34.57)(–0.74) by rounding each number to the nearest tenth. What was Jace’s estimate?

Mathematics

2 answers:

Naddika [18.5K]3 years ago

7 0

The answer is A. -25.6

RUDIKE [14]3 years ago

6 0

Answer:

jaces estimate

Step-by-step explanation:

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Distance between parallel lines y=3x+10 and y=3x-20

Alecsey [184]

1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).

2. Use formula $d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ to find the distance from point $(x_0,y_0)$ to the line Ax+By+C=0.

The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:

$d=\dfrac{|3\cdot 0-10-20|}{\sqrt{3^2+(-1)^2}}=\dfrac{30}{\sqrt{10}}=3\sqrt{10}$ .

3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.

Answer: $d=3\sqrt{10}$ .

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3 years ago

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Step-by-step explanation:

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2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

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A. Definition of Congruence

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What is the measure of each exterior angle for a regular polygon with 4 sides?

Yakvenalex [24]

A 90

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2 years ago

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