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3 years ago

11

The point p(0 3) is reflected across the y-axis. find the coordinates of its image

1 answer:

Vsevolod [243]3 years ago

7 0

A reflection on the y axis is (x,-y). So, the coordinates for this would be (0,-3).

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How do I simplify (6^-2)(3^-3)(3 x 6)^4? Not the answer, just the simplified expression.

Mazyrski [523]

The simplification form of the provided expression is 108 after applying the integer exponent properties.

<h3>What is integer exponent?</h3>

In mathematics, integer exponents are exponents that should be integers. It may be a positive or negative number. In this situation, the positive integer exponents determine the number of times the base number should be multiplied by itself.

We have an expression:

$=\left(6^{-2}\right)\left(3^{-3}\right)\left(3\:\cdot \:6\right)^4$

$=\dfrac{1}{36}\cdot \dfrac{1}{27}\left(3\cdot \:6\right)^4$

$=\dfrac{1}{36}\cdot \dfrac{1}{27}\cdot \:81\cdot \:1296$

$=\dfrac{1}{36}\cdot \dfrac{1}{27}\cdot \dfrac{104976}{1}$

$=\dfrac{104976}{972}$

= 108

Thus, the simplification form of the provided expression is 108 after applying the integer exponent properties.

Learn more about the integer exponent here:

brainly.com/question/4533599

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5 0

2 years ago

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

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3 years ago

Find the average rate of change for the interval from x=8 to x=9

Pepsi [2]

Answer:

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Step-by-step explanation:

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3 years ago

Continuity MS Grade 8 Pre-Algebra Qu. 4 (SY19-20) UPDATED

Tresset [83]

x: 3, 4, 6, 8, 9, 12

y: 2, 5, 7, 10, 11, 15

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4 years ago

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bonufazy [111]

Since there is one value of y for every value of x this is both a relation and a function

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4 years ago