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3 years ago

9

Point A (5, 6) is translated 3 units left and 5 units up, what are the coordinates of A prime?

1 answer:

Radda [10]3 years ago

3 0

Answer:

A' = (8, 11)

Explain:

5+3=8 & 6+5=11

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Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

<em><u>Option A</u></em>

<em><u>The solution is:</u></em>

$x = \pm 2i \sqrt{2}$

<em><u>Solution:</u></em>

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$

Thus the solution is found

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3 years ago

A bakery uses 550 pounds of sugar to make 1,000 cookies. Each cookie contains the same amount of sugar. How many pounds of sugar

Sonja [21]

Answer:

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Step-by-step explanation:

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3 years ago

The gravitational force exerted by the planet Earth on a unit mass at a distance r from the center of the planet is:

jolli1 [7]

Answer:

Yes, F is a continuous function of r

Step-by-step explanation:

We are given that

When r<R

$F(r)=\frac{GMr}{R^3}$

$r\geq R$

$F(r)=\frac{GM}{r^2}$

Where M=Mass of the earth

R=Radius of earth

G=Gravitational constant

We have to find the function is continuous of r or not.

LHL

$\lim_{r\rightarrow R-}\frac{GMr}{R^3}=\frac{GMR}{R^3}=\frac{GM}{R^2}$

RHL

$\lim_{r\rightarrow R+}\frac{GM}{R^2}=\frac{GM}{R^2}$

$F(R)=\frac{GM}{R^2}$

When a function is continuous at x=a

Then, LHL=RHL=f(a)

RHL==LHL=F(R)

Hence, the function is continuous of r.

8 0

3 years ago

Evaluate 3/7r + 5/8s when r = 14 and s = 8.

Ahat [919]

In this question, plug in 14 for r and 8 for s, which will look like 3/98 + 5/64. Solve, and you will get 0.109 = 0.11.

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3 years ago

What does Descartes' rule of signs tell you about the possible number of positive real zeros and the possible number of negativ

Sergio039 [100]

Answer:

3 or 1 positive real zeros
0 negative real zeros

Step-by-step explanation:

The signs of the coefficients of the given terms are ...

+ - + -

There are three sign changes, so the number of positive real zeros is 3 or 1.

When odd-degree terms have their signs changed, the signs become ...

- - - -

There are no sign changes, hence no negative real zeros.

_____

A graph confirms this evaluation.

8 0

3 years ago