Determine whether the two lines are parallel, perpendicular, or neither.
4x + 5y = 2
4x + 5y = 4

2.3+2.2y uneed to pus the 2 on the 3 and answer is uten

The distance that Greg is from home after 12 hours is given as follows:

300 miles.

To calculate the numeric value of a function or of an expression, we substitute each instance of any variable or unknown on the function by the value at which we want to find the numeric value of the function or of the expression presented in the context of a problem.

The function for this problem is given as follows:

y = 25x.

Hence the distance after 12 hours is given as follows:

y = 25 x 12

y = 300 miles.

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The exercise is designed to test the students knowledge of prepositions. The correct answer thus is (Option D) Yo estudio (which translates) "I study".

To understand the answer, you need to know what a preposition is. A preposition, in simple terms, is a word or group of words that comes before a noun.

The function of a preposition (much like an adjective) is to give clarity to the noun that it precedes.

Let us complete the Preposition phrase to give meaning to it:

"I study Mathematics". Where;

"Mathematics" is the noun;

"I Study" is the prepositional phrase. Hence the correction answer to the question above is Option D.

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Answer:

x = 2

Step-by-step explanation:

logx + log(2x - 1) = log 6

Apply log rules,

x(2x - 1) = 6

2x^2 - 1x = 6

x = 2 (true) or x = -3/2 (False)

Answer:

[tex]\dfrac a{b^2} + \dfrac b{a^2} = 10[/tex]

Step-by-step explanation:

[tex]\text{Given that, the roots are a,b and } ~ x^2 -4x+2 = 0\\\\\text{So,}\\\\a+b = -\dfrac{-4}1 = 4\\\\ab = \dfrac 21 = 2\\\\\text{Now,}\\\\~~~~~\dfrac a{b^2} + \dfrac b{a^2}\\\\\\=\dfrac{a^3 +b^3}{a^2b^2}\\\\\\=\dfrac{(a+b)^3 -3ab(a+b)}{(ab)^2}\\\\\\=\dfrac{4^3 -3(2)(4)}{2^2}\\\\\\=\dfrac{64-24}{4}\\\\\\=\dfrac{40}{4}\\\\\\=10[/tex]

Answer:

[tex]\dfrac{a}{b^2}+\dfrac{b}{a^2}=10[/tex]

Step-by-step explanation:

Given equation:   [tex]x^2-4x+2=0[/tex]

The roots of the given quadratic equation are the values of x when [tex]y=0[/tex].

To find the roots, use the quadratic formula:

[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]

Therefore:

[tex]a=1, \quad b=-4, \quad c=2[/tex]

[tex]\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}[/tex]

[tex]\textsf{Let }a=2+\sqrt{2}[/tex]

[tex]\textsf{Let }b=2-\sqrt{2}[/tex]

Therefore:

[tex]\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}[/tex]

Answer:

[tex]\frac{7\sqrt{65}}{65}[/tex]

Step-by-step explanation:

Cosine is the ratio of the side adjacent to the angle and the right triangle hypotenuse.

[tex]cos[/tex] B = [tex]\frac{7}{\sqrt{65}}[/tex] = [tex]\frac{7\sqrt{65}}{65}[/tex]

Answer:

9 children were jumping then.

Step-by-step explanation:

6+5 equals to 11 - 2 of the children left would equal 9 children left.

Answer:

66

Step-by-step explanation:

[tex]\sqrt{324} + 6 \sqrt{64}\\\\=\sqrt{18^2} +6 \sqrt{8^2}\\\\=18+6(8)\\\\=18+48\\\\=66[/tex]

Answer:

hmm

Step-by-step explanation:

not enough words to know u steal my points I steal yours kid

Answer:

y = 15

Step-by-step explanation:

Given :

3 + y = 18

This a simple algebraic equation.

Subtract 3 from both sides :

⇒ 3 + y - 3 = 18 - 3

⇒ y = 15

Answer:

y = 15

Step-by-step explanation:

Given equation:

To Find:

Solution:

We can rewrite this equation as:

[This'll ain't change the answer]

Now we could solve on a easy way.

STEPS:

Transpose +3 to the RHS, make sure to change it's sign from “+” to “-”.

=> y = 18-3

Subtract the integers which's on the RHS:

=> y = 15

Hence,the value of y will be 15.

[tex] \rule{225pt}{2pt}[/tex]

Answer:

V =113.09734

Step-by-step explanation:

[tex]V=\pi r^{2}[/tex]

[tex]= 2^{2}[/tex] x 9= 113.09734

(which you can technically round to 1)

=113.1

Solution :

[tex]x = \frac{4 + \sqrt{43} }{2} .x = 4 + \frac{4 - \sqrt{43} }{2} [/tex]

[tex] \frac{1}{(x - 5)} + \frac{3}{(x - 2)} - 4[/tex]

[tex]x = 2 + 3( x - 5) = 4 - (x - 5)(x + 2)[/tex]

[tex]x = \frac{4 + \sqrt{43} }{2} .x = 4 + \frac{4 - \sqrt{43} }{2} [/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{x-5}+\dfrac{3}{x+2}=4[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{x+2+3x-15}{(x-5)(x+2)}=4[/tex]

[tex]\\ \rm\Rrightarrow 4x-13=4(x-5)(x+2)[/tex]

[tex]\\ \rm\Rrightarrow 4x-13=4(x(x+2)-5(x+2))[/tex]

[tex]\\ \rm\Rrightarrow 4x-13=4(x^2+2x-5x-10)[/tex]

[tex]\\ \rm\Rrightarrow 4x-13=4(x^2-3x-10)[/tex]

[tex]\\ \rm\Rrightarrow 4x-13=4x^2-12x-40[/tex]

[tex]\\ \rm\Rrightarrow 4x^2-16x-53=0[/tex]

On solving we get

[tex]\\ \rm\Rrightarrow x=2\pm\dfrac{69}{2}[/tex]

Answer: 103/15

Step-by-step explanation:

We can simplify the right-hand side to be [tex]2 \left(\frac{1}{3}+6 \right)=2 \left(\frac{19}{3} \right)=\frac{38}{3}[/tex].

This means we need to solve:

[tex]\frac{1}{2}(5x-9)=\frac{38}{3}\\5x-9=\frac{76}{3}\\5x=\frac{103}{3}\\x=\boxed{\frac{103}{15}}[/tex]

Step-by-step explanation:

7x-5=30

Firstly -5 will cross the equality sign and become +5. so we'll have

7x=30+5=35

7x=35 we'll divide both sides by 7

giving u

7x/7=35/7

=x=5

therefore x=5.

CONFIRMATION

7*5-5=30

PLS MARK ME AS BRAINLIEST

Answer:

Step-by-step explanation:

Solve for X:

[tex](2,3) \longrightarrow (2+1, 3-3)=\boxed{(3,0)}\\(-4,5) \longrightarrow (-4+1, 5-3)=\boxed{(-3, 2)}\\(-3,4) \longrightarrow (-3+1, 4-3)=\boxed{(-2, 1)}[/tex]

Answer:

127.17 in³

Step-by-step explanation:

V = (1/2)πr²h
(1/2)(3.14)(2.25 in)²(16 in)

(1/2)(3.14)(5.0625 in²)(16 in) = 127.17 in³

The choice shows 19 + 32 + 11 rewritten correctly using the commutative property and then simplified correctly is option B; 19+11+32= 30+32 = 62.

The commutative property of addition says that it doesn't matter how we add two numbers, the result of the addition would be same.

For two numbers x and y, we have:

x + y = y + x

WE have given

19 + 32 + 11

The sum would be 62.

We know that by commutative property of addition,

For two numbers x and y,

x + y = y + x

Thus, if we take 3 numbers as a,b and c, then:

c + a + b = c + b + a

Similarly,

19 + 32 + 11 = 19+11+32

= 30+32

= 62

Therefore, the option B  is the correct answer.

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In function notation, f(x) is used instead of the letter y to represent the output variable.

A function is represented as:

f(x)

The above means that:

The function of x

As a general rule, the function can be rewritten using the letter y (i.e. the output variable)

Hence, the words that complete the blanks are y and output

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Answer: ......... 14.85

Step-by-step explanation:

4.95 × 3

...

14.85

bros lazy lol

Answer:

b) 4

Step-by-step explanation:

4 times $0.05 = $0.20

4 times $0.30 = $1.20

then you add the both totals together

0.20 + 1.20= $1.40

Answer:

(18,6)

Step-by-step explanation:

add 5 units to X and 2 to Y

Answer: is y=[x]-2

I used this app to help me get the answer too lol

a. If A is the coefficient matrix, solve det(A - λI) = 0 for the eigenvalues λ :

[tex]\det\begin{bmatrix}-6-\lambda & -4 \\ 12 & 8-\lambda\end{bmatrix} = (-6-\lambda)(8-\lambda)+48 = 0 \implies \lambda(\lambda-2)=0[/tex]

[tex]\implies \lambda = 0, \lambda = 2[/tex]

Let v = [v₁, v₂]ᵀ be the eigenvector corresponding to λ. Solve Av = λv for v :

[tex]\lambda=0 \implies \begin{bmatrix}-6&-4\\12&8\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \implies 3v_1 + 2v_2 = 0[/tex]

If we pick v₂ = -3, then v₁ = 2, so [2, -3]ᵀ is the eigenvector for λ = 0.

[tex]\lambda = 2 \implies \begin{bmatrix}-8&-4\\12&6\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \implies 2v_1 + v_2 = 0[/tex]

Let v₁ = 1, so v₂ = -2.

b. λ = 0 and v = [2, -3]ᵀ contributes a constant solution,

[tex]\vec y_1 = e^{\lambda t} v = \begin{bmatrix}2\\-3\end{bmatrix}[/tex]

while λ = 2 and v = [1, -2]ᵀ contribute a solution of the form

[tex]\vec y_2 = e^{\lambda t} v = e^{2t} \begin{bmatrix}1\\-2\end{bmatrix}[/tex]

c. Yes; compute the Wronskian of the two fundamental solutions:

[tex]W(1, e^{2t}) = \det\begin{bmatrix}1 & e^{2t} \\ 0 & 2e^{2t}\end{bmatrix} = 2e^{2t} \neq 0[/tex]

The Wronskian is non-zero, so the solutions are independent.

Answer:

20.5 - 5.36
= 15.14/2 (because we are using perimeter, so there are 2 equal lengths and 2 equal widths)
=7.57, therefore the width of the sandbox is 7.57 ft.

Answer:

4.89 ft

Step-by-step explanation:

Perimeter of a rectangle is given by the formula :

P = 2L + 2W

So now we solve for W :

P = 2(L+W)

P/2 = L+W

P/2 - L =  W

Now we substitute our P and L into the rearranged formula :

20.5 / 2 - 5.36 = W

10.25 - 5.36 = W

W = 4.89 ft

Hope this helped and brainliest please

Answer: Point C is (-2, -2), which is the third option.

Step-by-step explanation:

The midpoint formula is [tex]m=(\frac{x_{1}+x_{2} }{2},\frac{y_{1}+y_{2} }{2})[/tex]

All we have to do is plug in the numbers and solve the fractions from there.

[tex]m=(\frac{-10+6}{2},\frac{-6+2}{2} )[/tex]

[tex](\frac{-4}{2},\frac{-4}{2} )[/tex]

[tex]m=(-2, -2)[/tex]

I hope this helps!!

Substitute -2 for x

[tex]f( - 2) = 2( - 2) + 3 \\ y = - 4 + 3 \\ y = - 1[/tex]

Hope it helps

Please give brainliest

Answer:

-1

Step-by-step explanation:

Hi student! Let me help you out on this question.

_____________________

To find the value of f(-2), we need to stick in -2 for x.

[tex]\mathsf{f(-2)=2\cdot(-2)+3}[/tex]. Multiply first.

[tex]\mathrm{f(-2)=-4+3}[/tex]. Now simplify completely.

[tex]\mathsf{f(-2)=-1}[/tex]. Which is our final answer.

Hope that this helped you out! have a good day ahead.

Best Wishes!

[tex]\star\bigstar\underline{\underline{\overline{\overline{\textsf{Reach far. Aim high. Dream big.}}}}}\bigstar\star[/tex]

◆◈-Greetings!-◆◈

__________________

[tex](\stackrel{x_1}{-5}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{7}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{7}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{-2}-\underset{x_1}{(-5)}}} \implies \cfrac{7 -1}{-2 +5}\implies \cfrac{6}{3}\implies 2[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{2}(x-\stackrel{x_1}{(-5)})\implies y-1=2(x+5)[/tex]

The angles are 121 and 59 degrees, respectively

The given parameters are:

FGH = 11x

HGJ = 6x - 7

Linear pair angles add up to 180.

So, we have:

11x + 6x - 7 = 180

Evaluate the like terms

17x = 187

Divide both sides by 17

x = 11

Substitute x = 11 in FGH = 11x and HGJ = 6x - 7

FGH = 11 * 11=121

HGJ = 6 * 11 - 7 = 59

Hence, the angles are 121 and 59 degrees, respectively

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