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3 years ago

Need some help ya plz

Mathematics

2 answers:

Alex73 [517]3 years ago

6 0

Answer:

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y )

Hence

C(5, 3 ) → C'(- 5, 3 ) → C

erastovalidia [21]3 years ago

3 0

Answer:

C. (-5, 3)

Step-by-step explanation:

Point C is currently (5,3).

If it is reflected over the y-axis, that means while the <em>y-coordinate stays the same, the x- coordinate is negative of what is was before</em>.

C is the only answer that keeps the y-coordinate constant, so just by looking at that, we can say that the answer is C.

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HELP. How long is the minor axis for the ellipse shown below?

kifflom [539]

Given:

The equation of ellipse is

$\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1$

To find:

The length of the minor axis.

Solution:

The standard form of an ellipse is

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ ...(i)

where, (h,k) is center, if a>b, then 2a is length of major axis and 2b is length of minor axis.

We have,

$\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1$ ...(ii)

On comparing (i) and (ii), we get

$b^2=16$

Taking square root on both sides.

$b=\pm 4$

Consider only positive value of b because length cannot be negative.

$b=4$

Now,

Length of minor axis = $2b$

= $2(4)$

= $8$

So, the length of minor axis is 8 units.

Therefore, the correct option is B.

8 0

3 years ago

Simplify the expression pls :(

gulaghasi [49]

The answer is 68 because 11 - 3 is 8 and 8 squares is 64 and 64 + 4 is 68.

7 0

3 years ago

Hello please help me solve this :) will give braisnlt

Serjik [45]

I wrote ones but I hope it’s right and helps

3 0

3 years ago

Read 2 more answers

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago

I really need help plz ok

Kryger [21]

Answer:

n = 12

Ava has 12 times as many eggs as Cole

Step-by-step explanation:

May I have brainliest please? :)

4 0

3 years ago

Read 2 more answers