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2 years ago

Janelle practices basketball every

Mathematics

1 answer:

enyata [817]2 years ago

3 0

Janelle practices every afternoon

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Help solve for r algebra

bija089 [108]

Answer:

$\frac{20}{7}$

Step-by-step explanation:

Multiply both sides by 7 to get rid of the denominator on the Right Side (RS). Do the same for the left side (LS) by multiplying both sides by 5 to get:

$7r=4(5)$

Simplify: $7r=20$

Divide both sides of the equation by 7 to get: $r=\frac{20}{7}$

6 0

3 years ago

Identify the ellipses, represented by equations, whose eccentricities are less than 0.5

ozzi

Answer:

See explanation

Step-by-step explanation:

Hello, we cannot see the ellipse equations.

The eccentric of an ellipse is given by:

$e = \frac{c}{a}$

Assuming the equations are:

$\frac{ {x}^{2} }{25} + \frac{ {y}^{2} }{16} = 1$

Then a²=25 and b²=16

${c}^{2} = {a}^{2} - {b}^{2}$

${c}^{2} = 25 - 16$

${c}^{2} = {3}^{2}$

$c = 3$

The eccentricity is:

$e = \frac{3}{5} = 0.6$

If the ellipse has equation,

$\frac{ {x}^{2} }{ {5}^{2} } + \frac{ {y}^{2} }{ {3}^{2} } = 1$

then the this time, we have a=5 and b=3.

This means that:

${c}^{2} = {a}^{2} - {b}^{2}$

${c}^{2} = {5}^{2} - {3}^{2}$

${c}^{2} = {4}^{2}$

$c = 4$

$c = \frac{4}{5} = 0.8$

3 0

2 years ago

Read 2 more answers

(Divide the following polynomials (x5 + y5) ÷ (x + y)

stiks02 [169]

$\text{Use}\ a^5+b^5=(a + b) (a^4 - a^3 b + a^2 b^2 - a b^3 + b^4).\\\\(x^5+y^5):(x+y)=\dfrac{x^5+y^5}{x+y}=\dfrac{(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)}{x+y}\\\\=\boxed{x^4-x^3y+x^2y^2-xy^3+y^4}$

7 0

3 years ago

BRAINLIEST AND 50 PTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!HURRY!!!!!!!!!!!!!

Lana71 [14]

Answer:

Step-by-step explanation

$(8 * 10) + ( 2 * \frac{1}{100} ) + ( 9 * \frac{1}{1000} )$

⇒ 80 + ( 2 * 0.01 ) + ( 9 * 0.001 )

⇒ 80 + 0.02 + 0.009

= 80.029

6 0

3 years ago

The three medals earned in the Olympics are the gold, silver, and bronze medals. In the 2014 Winter Olympics, the United States

marta [7]

Answer:

Let x be the number of silver medals.

As there were two more gold medals than silver ones, gold medals are x+2

We also know that the number of bronze medals was 4 less than the sum of gold and silver, so if there are x + 2 of gold and x of silver, there are x+x+2-4 of bronze.

Now, we can do an equation, as we know there were a total of 28 medals:

x + x + 2 + x + x + 2 - 4 = 28

And we isolate x:

4x = 28

x = 28/4 = 7

There were 7 silver medals, so there were 9 gold ones (7-2) and 12 of bronze (9+7-4).

4 0

2 years ago