If -5x-y=1 is a true equation, what would be the value of 5(-5x-y)?

Write 90% as a decimal and fraction in simplest form

solong [7]

90% as a decimal is .9 and then as a fraction in simplest form is 9/10
Hope it helps

5 0

3 years ago

Verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3. Also verify the relationship between the ze

Finger [1]

Solution:

we have been asked to verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial $4x^3+20x^2+2x-3$

To verify that whether the given values are zeros or not we will substitute the values in the given Polynomial, if it will returns zero, it mean that value is Zero of the polynomial. But if it return any thing other than zeros it mean that value is not the zero of the polynomial.

Let $f(x)=4x^3+20x^2+2x-3\\\\\text{when x=-5}\\\\f(-5)=4(-5)^3+20(-5)^2+2(-5)-3=-13\\\\\text{when x=}\frac{1}{2}\\$

$f( \frac{1}{2} ) = 4 ( \frac{1}{2} )^3+20(\frac{1}{2})^2+2(\frac{1}{2})-3=\frac{7}{2}\\\\$

$\text{when x=}\frac{3}{4}\\\\$

$f( \frac{3}{4} ) = 4 ( \frac{3}{4} )^3+20(\frac{3}{4})^2+2(\frac{3}{4})-3=\frac{183}{16}\\$

Hence -5, 1/2, and 3/4 are not the zeroes of the given Polynomial.

Since sum of roots $=\frac{-b}{a}= \frac{-20}{4}=-5\\$

But $-5+\frac{1}{2}+\frac{3}{4}= \frac{-15}{4}\neq-5$

Hence we do not find any relation between the coefficients and zeros.

Anyway if the given values doesn't represents the zeros then those given values will not have any relation with the coefficients of the p[polynomial.

6 0

3 years ago

Peter wants to purchase pizza pies and breadsticks for a party. The cashier tells him that pizza pies are $8 each and breadstick

sladkih [1.3K]

The answer is 5x+8y<= 120

6 0

3 years ago

Read 2 more answers

Bianca filled 2 same-sized jars with flavored popcorn as a gift. She wants to glue a piece of ribbon around the edge of each lid

dybincka [34]

I think it is 177.4728

but im not sure let me check

7 0

3 years ago

Read 2 more answers

The planets in our solar system do not travel in circular paths. Rather, their orbits are elliptical. The Sun is located at a fo

qwelly [4]

1. The distance between the perihelion and the aphelion is 116 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Step-by-step explanation:

Let us revise the equation of the ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where the major axis is parallel to the x-axis

The length of the major axis is 2a
The coordinates of the vertices are (± a , 0)
The coordinates of the foci are (± c , 0) , where c² = a² - b²

∵ The Sun is located at a focus of the ellipse

∴ The sun located ate c

∵ The perihelion is the point in a planet’s orbit that is closest to the

Sun ( it is the endpoint of the major axis that is closest to the Sun )

∴ The perihelion is located at the vertex (a , 0)

∵ The closest Mercury comes to the Sun is about 46 million miles

∴ The distance between a and c is 46 million miles

∵ The aphelion is the point in the planet’s orbit that is furthest from

the Sun ( it is the endpoint of the major axis that is furthest from

the Sun )

∴ The aphelion is located at the vertex (-a , 0)

∵ The farthest Mercury travels from the Sun is about 70 million miles

∴ The distance from -a to c is 70 million miles

∴ The distance between the perihelion and the aphelion =

70 + 46 = 116 million miles

1. The distance between the perihelion and the aphelion is 116 million miles

∵ The distance between the perihelion and the aphelion is the

length of the major axis of the ellipse

∵ The length of the major axis is 2 a

∴ 2 a = 116

- Divide both sides by 2

∴ a = 58

∴ The distance from the center of Mercury’s elliptical orbit to the

closest end point to the sun is 58 million miles

∵ The distance between the sun and the closest endpoint is

46 million miles

∴ The distance from the center of Mercury’s elliptical orbit and

the Sun = 58 - 46 = 12 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

∵ The major axis runs horizontally

∴ The equation is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

∵ a = 58

∵ c is the distance from the center to the focus of the ellipse

∴ c = 12

∵ c² = a² - b²

∴ (12)² = (58)² - b²

- Add b² to both sides

∴ (12)² + b² = (58)²

- Subtract (12)² from both sides

∴ b² = (58)² - (12)² = 3220

- Substitute these values in the equation

∴ $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

The eccentricity (e) of an ellipse is the ratio of the distance from the

center to the foci (c) and the distance from the center to the

vertices (a) ⇒ $e=\frac{c}{a}$

∵ c = 12

∵ a = 58

∴ $e=\frac{12}{58}$ = 0.207

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

If the eccentricity is zero, it is not squashed at all and so remains a circle.

If it is 1, it is completely squashed and looks like a line

∵ The eccentricity of the ellipse is 0.207

∵ This number is closed to zero than 1

∴ The shape of the ellipse is near to the shape of the circle

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Learn more:

You can learn more about conics section in brainly.com/question/4054269

#LearnwithBrainly

5 0

3 years ago