All categories

3 years ago

BEST ANSWER WILL BE MARKED BRAINIEST W/ EXPLANATION!!!

Mathematics

1 answer:

natka813 [3]3 years ago

7 0

D₀,₃ (x , y), where:

D stands for dilatation. the sub (0) = center and the sub (3) represents the scale factor. The scale factor "k" being 3, means the actual image has been dilated 3 times versus the pre-image or reciprocally, the pre-image is a reduction of the image by a factor k₁= 1/3

The coordinate of A' in the image A(-3, 6)
The coordinate of A in the pre-image should be A(-3.k₁, 6k₁)

But k₁ =1/3, then A(-3/3 , 6/3) = A(-1 , 2) (Second Answer)

You might be interested in

Convert to liters.<br><br> 3,900 mililiters

Klio2033 [76]

It’s 3.9 because if you write out the metric system K H D L D C M. From milliliters to liters is three spaces so for the num. 3900 go backwards three spaces

5 0

3 years ago

1. Find the slope and y-intercept of the line.

Fynjy0 [20]

Answer:

i do not know

Step-by-step explanation:

well first off u didn't give the graph at all which makes it impossible to answer.

8 0

3 years ago

1/2 X 2/3 multiply fraction

kicyunya [14]

Answer:

If you take 2/3 in a calculator you get 0.6 repeating and if you take that times 1/2 you will get 0.3 repeating therefor 1/3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Use synthetic division and the Remainder Theorem to find P(a). P(x) = 2x3 + 4x2 − 10x − 9; a = 3

guapka [62]

Answer:

Step-by-step explanation:

Use synthetic division and the Remainder Theorem to find P(a). P(x) = 2x3 + 4x2 − 10x − 9; a = 3

8 0

3 years ago

t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.

kondaur [170]

Answer:

a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

7 0

4 years ago