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IrinaVladis [17]

3 years ago

6

Which of the following equations represents a horizontal line through (5,-3)?

2 answers:

Novay_Z [31]3 years ago

4 0

it shows that y=5 is correct

Snowcat [4.5K]3 years ago

3 0

X=5 vertical all the best

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Help sense No one understands.

Blababa [14]

I think either 3 or 7

7 0

3 years ago

Read 2 more answers

William has 24 24 cans of fruit and 60 60 cans of vegetables that he will be putting into bags for a food drive. He wants each b

Snezhnost [94]

Answer:

William will make 12 bags of food and each of the bag will contains 2 cans of fruit and 5 cans of vegetables.

Step-by-step explanation:

Given:

Number of fruits cans = 24

Number of veggies cans = 60

William will have to distribute them in equal bags with equal cans of fruits and vegetables respectively.

For this:

We have to find the GCF (greatest common factor) of 24 and 60.

GCF by listing out the factors method.

Factors of 24 : $1,2,3,4,6,8,12,24$

Factors of 60 : $1,2,3,4,5,6,10,12,15,20,30,60$

So,

The greatest common factor of $24$ and $60$ is $12$ .

The number of bags William will used for equal distribution = $12$

Now,

We have to distribute the veggies and fruits in equal number of cans to these $12$ bags.

Number of fruits cans used in each bag = $\frac{24}{12} = 2$

Number of vegetables can used in each bag = $\frac{60}{12} =5$

We can say that:

William will make 12 bags of food and each of the bag will contains 2 cans of fruit and 5 cans of vegetables.

4 0

3 years ago

Do you agree with the statement that " cubic functions must have at least 1 x-intercept but not more than 3, whereas quadratic f

vladimir2022 [97]

Answer:

Yes, I agree

Step-by-step explanation:

For the cubic function

A cubic function is represented as:

$f(x)=ax^3+bx^2+cx+d$

A cubic function may have 1, 2 or 3 x intercepts. This is shown below

For 3 x intercepts

$y = x^3 - x$

Equate y to 0

$x^3 - x = 0$

Expand

$x(x^2 - 1) = 0$

Express $x^2 - 1$ as difference of two squares

$x(x - 1)(x +1 ) = 0$

<em>x = 0 or 1 or -1</em>

For 2 x intercepts

$y = x^3 - x$

$y =(x-5)^2)(x+7)$

Equate y to 0

$(x-5)^2(x+7) = 0$

Expand

$(x-5)(x-5)(x+7) = 0$

<em>x= 5 or x = -7</em>

For 1 x intercept

$y = x^3$

Equate y to 0

$x^3 = 0$

Take cube roots of both sides

$x = 0$

It has been shown above that a cubic function may have 1, 2 or 3.

So, I agree to the statement

For the quadratic function

A quadratic function will not have any x intercept when the function can not be factorized;

E.g.

$y = x^2 + x + 17$

<em>The above function has no x intercept.</em>

A quadratic function will have at least 1 x intercept when the function can be factorized;

E.g.

$y = x^2- 6x + 9$

Equate y to 0

$x^2- 6x + 9 = 0$

Expand

$x^2 - 3x - 3x + 9 = 0$

$(x - 3)(x-3) = 0$

$x = 3$

We've shown that a quadratic may have no x intercept, and it may also have x intercept(s)

Hence, I agree to both statement

3 0

2 years ago

Which voting methods satisfy the Majority Fairness Criterion?

leonid [27]

It is a single-winner voting system criterion used to compare such systems.

3 0

3 years ago

Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.

natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is $y=x-x^2-2$ and the solutions are defined for all real numbers.

Step-by-step explanation:

The given differential equation is

$y'=1-2x$

It can be written as

$\frac{dy}{dx}=1-2x$

Use variable separable method to solve this differential equation.

$dy=(1-2x)dx$

Integrate both the sides.

$\int dy=\int (1-2x)dx$

$y=x-2(\frac{x^2}{2})+C$ $[\because \int x^n=\frac{x^{n+1}}{n+1}]$

$y=x-x^2+C$ ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

$-2=1-(1)^2+C$

$-2=1-1+C$

$-2=C$

The value of C is -2. Substitute C=-2 in equation (1).

$y=x-x^2-2$

Therefore the solution of the given initial value problems in explicit form is $y=x-x^2-2$ .

The solution is quadratic function, so it is defined for all real values.

Therefore the solutions are defined for all real numbers.

4 0

3 years ago