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2 years ago

Please help me with math please please

Mathematics

1 answer:

Law Incorporation [45]2 years ago

6 0

Answer:

subtract 42 - 7

Step-by-step explanation: u always solve the parenthesis first

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Which number is not equivalent to 467%?

Burka [1]

It could be almost anything, but an example is three (3).

8 0

2 years ago

CAN SOMEONE PLS HELP ME!

kolbaska11 [484]

1ft 9 in= 21 inches.

6ft=72 inches.

72inches - 21inches = 51 inches.

The hole needs to be 51 inches deeper

4 0

1 year ago

Find the center, vertices, and foci of the ellipse with equation 4x^2 + 9y^2 =36

Anvisha [2.4K]

<u>Answer: </u>

The center, vertices and foci of the ellipse with equation $4 x^{2}+9 y^{2}=36 is (0,0),(\pm 3,0),(\pm \sqrt{5}, 0)$ respectively

<u> Solution: </u>

The equation of ellipse with centre (0, 0) in the form of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ --- eqn 1

Where,

x is the major axis

Centre (0, 0)

Vertices is $(\pm \mathrm{a}, 0)$

Foci is $(\pm \mathrm{c}, 0)$ where $c=\sqrt{a^{2}-b^{2}}$

Now given that the equation of ellipse is

$4 x^{2}+9 y^{2}=36$ --- eqn 2

On dividing equation (2) by 36,

$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

On comparing equations (1) and (2),

We get a = 3, b= 2

$c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}$

So centre of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (0, 0)$

Vertices of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm 3,0)$

Foci of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm \sqrt{5}, 0)$

7 0

3 years ago

Hello im in school now i would be less active so yea lol

Kruka [31]

Answer:

I am an alien. I like playing basketball and football.

Step-by-step explanation:

6 0

2 years ago

Evaluate the following expression when x = 6 and y = 2:

Yuri [45]

X=6
y=2

(x^2+y^3)/(2+x)
(6^2+2^3)/(2+6)
(36+8)/(2+6)
44/8
5 1/2

Hope this helps :)

3 0

2 years ago

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