Graph the system of equations on graph paper to answer the question.
y =2/7x-7
y=-x+ 2
What is the solution for this system of equations?
Enter your answer in the boxes. HELPP

Answer:

A. median

explain i just aced the test rn

Answer:

Angle bisector

Step-by-step explanation:

The line AD bisects both sides of the triangle. Both angles are congruent to each other since both sides are congruent to each other.

[tex]~~~~~~~\log_{10} N = 3.8609\\\\\\\implies N = 10^{3.8609} ~~~~~~~~~;[\log_b c =d \implies b^d = c]\\\\\\\implies N \approx 7259[/tex]

Answer:

They are both right-angled triangles.

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

height (y-axis) = constant × width (x-axis)

let's look at the point coordinates.

remember, an ordered pair of point coordinates consists of an x value and an associated (typically calculated by a function) y value.

we see that all y values are half of the x values (3 and 1.5, 2 and 1, 7 and 3.5, 8 and 4).

so, the constant in our equation is responsible for returning half of the x value back as the y value.

what constant factor turns a number in half ?

x × c = x/2

c = 1/2

and so, A is the right answer.

Answer:

[tex]4\sqrt[3]{307}[/tex] o 26.98398684 en forma decimal

Step-by-step explanation:

For the given function, the domain is D : { x ≥ 7} and the range is R: { y ≥ 9}

Here we have a square root, remember that the argument of the square root must be equal or larger than zero, so the domain is such that:

x - 7 ≥ 0.

Solving for x we get:

x ≥ 0 + 7

Then the domain is:

x ≥ 7

To get the range, we evaluate in the minimum of the domain:

f(7) = √(7 - 7) + 9 = 9

Then the range is the set of all values larger than 9, because the function is increasing.

So the range is R: y ≥ 9.

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Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

[tex]\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}[/tex]

Cramer's rule says that

[tex]x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}[/tex]

where [tex]x_i[/tex] is the solution for i-th variable, and [tex]\mathbf A_i[/tex] is a modified version of [tex]\mathbf A[/tex] with its i-th column replaced by [tex]\mathbf b[/tex].

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

[tex]\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}[/tex]

[tex]\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}[/tex]

[tex]\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))[/tex]

[tex]\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120[/tex]

The modified matrices and their determinants are

[tex]\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240[/tex]

[tex]\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360[/tex]

[tex]\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480[/tex]

Then by Cramer's rule, the solution to the system is

[tex]x = \dfrac{-240}{-120} \implies \boxed{x = 2}[/tex]

[tex]y = \dfrac{360}{-120} \implies \boxed{y = -3}[/tex]

[tex]z = \dfrac{-480}{-120} \implies \boxed{z = 4}[/tex]

Answer:

in photo attached.

Step-by-step explanation:

By quadratic formula, the roots of the quadratic function 2 · x² - 2 · x + 5 = 0 are two conjugated complex numbers: x₁ = 0.5 + i 1.5 and x₂ = 0.5 - i 1.5, respectively.

Let be a quadratic function of the form a · x² + b · x + c = 0, whose roots can be found by means of the following formula:

[tex]x = \frac{-b \pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}[/tex]     (1)

Where a, b, c are the coefficients of the quadratic function.

In we know that 2 · x² - 2 · x + 5 = 0, then the roots of the polynomial are, respectively:

[tex]x_{1} = \frac{2 + \sqrt{(-2)^{2}-4\cdot (2)\cdot (5)}}{2\cdot (2)}[/tex]

[tex]x_{1} = \frac{2 + \sqrt{4-40}}{4}[/tex]

x₁ = 0.5 + i 1.5

[tex]x_{2} = \frac{2 - \sqrt{(-2)^{2}-4\cdot (2)\cdot (5)}}{2\cdot (2)}[/tex]

[tex]x_{2} = \frac{2 - \sqrt{4-40}}{4}[/tex]

x₂ = 0.5 - i 1.5

By quadratic formula, the roots of the quadratic function 2 · x² - 2 · x + 5 = 0 are two conjugated complex numbers: x₁ = 0.5 + i 1.5 and x₂ = 0.5 - i 1.5, respectively.

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Answer: The correct answers are the first and the fourth one.

Step-by-step explanation:

The exponential function of the table is y = 0.75(2)^x

From the table, we have the following points

(x,y) = (0,0.75) and (1,1.5)

An exponential function is represented as:

y = ab^x

At (0,0.75), we have:

0.75 = ab^0

a = 0.75

Substitute a = 0.75 in y = ab^x

y = 0.75b^x

At (1,1.5), we have:

1.5 = 0.75b^1

Evaluate

1.5 = 0.75b

Divide both sides by 0.75

b =2

Substitute b =2 in y = 0.75b^x

y = 0.75(2)^x

Hence, the exponential function is y = 0.75(2)^x

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The system of equations have one solution

The equations are given as:

2x + 2y = 5

3x + 3y = 7

The above equations are distinct linear equations.

This means that they would have one point of intersection, if plotted on a graph

Hence, the system of equations have one solution

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Based on the calculations, the first three (3) iterations of the given function are 120, 96 and 76.8.

In this exercise, you're required to find the first three (3) iterations of the given function. Thus, we would substitute the value of x₀ into the function and then evaluate as follows:

First iteration:

f(x) = 0.80x

f(x₀) = 0.80x₀

f(150) = 0.80 × 150

f(150) = 120.

Second iteration:

f(x) = 0.80x

f(x₁) = 0.80x₁

f(120) = 0.80 × 120

f(120) = 96.

Third iteration:

f(x) = 0.80x

f(x₂) = 0.80x₂

f(96) = 0.80 × 96

f(120) = 76.8.

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Answer:

y = 7x +2

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (0, 2 ) and (x₂, y₂ ) = (- 1, - 5 )

m = [tex]\frac{-5-2}{-1-0}[/tex] = [tex]\frac{-7}{-1}[/tex] = 7

the line crosses the y- axis at (0, 2 ) ⇒ c = 2

y = 7x + 2 ← equation of line

[tex]~~~~~~k = \dfrac{3n+2}{n+1}\\\\\implies k(n+1) = 3n+2\\\\\implies kn+k=3n+2\\\\\implies kn -3n= 2-k\\\\\implies n(k-3) = 2-k\\\\\implies n = \dfrac{2-k}{k-3}\\\\\implies n = -\dfrac{k-2}{k-3}~~~~~~~~;[k\neq 3][/tex]

Answer:

-3y+8z

Step-by-step explanation:

2x-2y+5z-2x-y+3z

you don't need to change their signs just place them accordingly

2x-2x-2y-y+5z+3z

-3y+8z

The ratio between the perimeter of the largest and smallest pentagon is 2.

We know that the pentagons are similar, meaning that the dimensions of one of the pentagons is k times the dimensions of the other.

Because of this, the ratio between the areas is k squared. And because the perimeter depends linearly on the dimensions, the ratio between the perimeters will be equal to k.

So we need to find k, we will have:

[tex]\frac{144 in^2}{36 in^2} = k^2 = 4\\\\k = \sqrt{4} = 2[/tex]

Then we conclude that the ratio between the perimeters is k  =2.

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This exercise is about creating two-dimensional shapes. The resulting shape is a quadrilateral - Square. See the attached for the lines drawn.

What was noticed about the two lines drawn?

When two lines intersect with one another such that they create a right angle, perpendicularity has occurred and both lines are said to be perpendicular to one another.

Hence:

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The distribution of the values obtained from a simple random sample of size n from the same population is incorrect.

The sampling distribution of a statistic of size n is the distribution of the values obtained from a simple random sample of size n from the same population.

The sampling distribution is the process of getting a sample through simple random techniques from the sample population.

So, it is incorrect that the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population.

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Answer:

simple interest

Step-by-step explanation:

SI = principal x rate x time

Using the combination formula, it is found that:

There are 364 different orders of top-three finishers that include all seniors.

The probability that the top-three finishers will all be seniors is 15.38%.

The order in which the players are taken is not important, hence the combination formula is used to solve this question.

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In total, three students are taken from a set of 14, hence:

[tex]C_{14,3} = \frac{14!}{3!11!} = 364[/tex]

Including only seniors, it would be three students from a set of 8, hence:

[tex]C_{8,3} = \frac{8!}{3!5!} = 56[/tex]

Hence the probability is given by:

p = 56/364 = 0.1538 = 15.38%.

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The probability that the student selected at random is a freshman is; 29%

From the given table;

Total number of male students = 4 + 6 + 2 + 2 = 14

Number of freshmen students = 4 students

Thus;

Probability that the student selected at random is a freshman is;

P(Freshman | Male) = 4/14 * 100% = 28.57% ≈ 29%

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Answer:

3/7

Step-by-step explanation:

Probability=Number of possible items÷Number of total items.

We will make the possible item 3 because we are looking for the probability whether a boy will be picked.

The number of total items will therefore be: the number of boys + the number of girls; that will be 3+4 =7.

So the probability of picking a boy will be 3/7

THANK YOU.

Answer:

uh I'm not so sure but

Step-by-step explanation:

it might be 1664

Answer:

The first step for deriving the quadratic formula from the quadratic equation, 0 = ax2 + bx + c, is shown.

[tex] \: first the formula of quadratic equation \: = - b ≠ \: \sqrt{ \frac{ \: {b }^{2} - 4ac}{ \: 2a} } [/tex]

let's begin to solve this equation,,,

[tex]ax² + bx + c = 0 \\

ax² + bx + c =0 \\ {x}^{2} + \frac{b}{a} \: + \frac{ {b}^{2} }{2a} = \frac{b2}{a} - \frac{c}{a}

\\ x + \frac{ {b}^{2} }{2a} = - \frac{ {b}^{2} }{2a} - \frac{c}{a} [/tex]

an equation having the maximum power of the variable equal to is called the quadratic equation the general form of quadratic equation is ax² + bx +c =0 where a,b,c are real numbers a ≠0 x is variable •

a quadratic equation can be solved by two method by factorization and by formula

by factorization a quadratic equation can be solved by factorization only when the product AC can be divided into two such part that it had the sum of the difference of the two part is equal to b

[tex]by the formula of quadratic equation can be solved by \ - } [/tex]

[tex]x = \frac{ - b \sqrt{ {b}^{2} - 4ac } }{2a } \\ root \: of \: equation \: {ax}^{2} + bx + c = 0 \\ - b - \sqrt{ \frac{ {b}^{2} - 4ac}{2a} } [/tex]

Answer:

Molly's jump rope is 1 2/3 feet longer than Gail's jump rope.

Step-by-step explanation:

To get this answer, you would have to subtract 6 1/3 - 4 2/3 to find the difference between the two lengths.

To solve this subtraction problem: Start by doing 6 - 4 to get 2, because when the fraction parts are like that, start with whole numbers. Then, solve the fraction parts. 1/3 - 2/3 is -1/3 (you can go to negatives)! Lastly, combine the whole number and fraction parts. 2 - 1/3, because the negative symbol changes to a minus symbol, which will get you 1 2/3.

Hope this helped!

The distance of the robot from the charging base is gotten as; 62 cm

From the image attached showing the movement of Kapil's robot, we can use cosine rule to find the value of h which is the distance of the robot from the charging base.

The distance of the robot from the charging base is gotten by;

h = x² + b² - 2xb cos 60

h = 50² + 70² - 2*50*70 * 0.5

h = 62 cm

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