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4 years ago

7

True or False: Tangent line approximation works at any point on any graph.

1 answer:

Tresset [83]4 years ago

5 0

That is false. Tangent line has to start from a specific line( either the (x-axis) or the (y-axis).

Hope this helped. :)

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Find the equation of the line that passes through the points (-3,0) and (4,2). Using point slope formula

astra-53 [7]

Answer:

$\large\boxed{y-0=\dfrac{2}{7}(x+3)}\\\downarrow\\\boxed{y=\dfrac{2}{7}x+\dfrac{6}{7}}$

Step-by-step explanation:

The point-slope form of an equation of a line:

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

<em>m</em><em> - slopei</em>

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

======================================

We have the points (-3, 0) and (4, 2).

Substitute:

$m=\dfrac{2-0}{4-(-3)}=\dfrac{2}{7}$

Put the value of the slope and the coordinates of the point (-3, 0) to the equation:

$y-0=\dfrac{2}{7}(x-(-3))$

$y-0=\dfrac{2}{7}(x+3)$

Converto to the slope-intercept form (<em>y = mx + b</em>):

$y=\dfrac{2}{7}x+\dfrac{6}{7}$

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4 years ago

Jada is evaluating 8 × 3 × 4. She doesn't know the product of 24 and 4 off the top of her head, so she decides to multiply 3 by

siniylev [52]

Answer:

plz explain correctly please

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3 years ago

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Last month sue made $200 by baby-sitting for 12 hours and tutoring for 20 hours. This month she worked the same number of hours

stiks02 [169]

$44, she needs to step up her game

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3 years ago

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What is 7 times 7 plus 7

mr Goodwill [35]

Answer:

7×7+7

49+7

=56

order of operations multiply first then add

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3 years ago

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Show that W is a subspace of R^3.

musickatia [10]

Answer:

Check the two conditions of Subspace.

Step-by-step explanation:

If W is a Subspace of a vector space, V then it should satisft the following conditions.

1) The zero element should be in W.

Zero element can be different for different vector spaces. For examples, zero vector in $\math{R^2}$ is (0, 0) whereas, zero element in $\math{R^3}$ is (0, 0 ,0).

2) For any two vectors, $w_1$ and $w_2$ in W, $w_1 + w_2$ should also be in W.

That is, it should be closed under addition.

3) For any vector $w_1$ in W and for any scalar, $k$ in V, $kw_1$ should be in W.

That is it should be closed in scalar multiplication.

The conditions are mathematically represented as follows:

1) 0 $\in$ W.

2) If $w_1 \in W; w_2 \in W$ then $w_1 + w_2 \in W$ .

3) $$ \forall k \in V, and \hspace{2mm} \forall w_1 \in W \implies kw_1 \in W$

Here V = $\math{R^3}$ and W = Set of all (x, y, z) such that $x - 2y + 5z = 0$

We check for the conditions one by one.

1) The zero vector belongs to the subspace, W. Because (0, 0, 0) satisfies the given equation.

i.e., 0 - 2(0) + 5(0) = 0

2) Let us assume $w_1 = (x_1, y_1, z_1)$ and $w_2 = (x_2, y_2, z_2)$ are in W.

That means: $x_1 - 2y_1 + 5z_1 = 0$ and

$x_2 - 2y_2 + 5z_2 = 0$

We should check if the vectors are closed under addition.

Adding the two vectors we get:

$w_1 + w_2 = x_1 + x_2 - 2(y_1 + y_2) + 5(z_1 + z_2)$

$= x_1 + x_2 - 2y_1 - 2y_2 + 5z_1 + 5z_2$

Rearranging these terms we get:

$x_1 - 2y_1 + 5z_1 + x_2 - 2y_2 + 5z_2$

So, the equation becomes, 0 + 0 = 0

So, it s closed under addition.

3) Let k be any scalar in V. And $w_1 = (x, y, z) \in W$

This means $x - 2y + 5z = 0$

$kw_1 = kx - 2ky + 5kz$

Taking k common outside, we get:

$kw_1 = k(x - 2y + 5z) = 0$

The equation becomes k(0) = 0.

So, it is closed under scalar multiplication.

Hence, W is a subspace of $\math{R^3}$ .

7 0

3 years ago