solve in systems by substitution y=x+2 and 3x+3y=6

Answer:

1:6

Step-by-step explanation:

5:30 simplified is 1:6

Answer:

5:30 Simplified: 1:6

Step-by-step explanation:

Hope this helps, and don't forget to follow!

Answer:

C

Step-by-step explanation:

Substitute t = 5.5 into h(t) , that is

h(5.5) = - 16(5.5)² + 128(5.5) = - 484 + 704 = 220 ft → C

Answer:

i think 6 is ans

hope u like it

Answer:

Any 3 trinalges

Step-by-step explanation:

total triangles= 9

1/3 of 9 = 1/3 x 9 =  3

So, shade 3 triangles

I hope im right!1

Step-by-step explanation:

There are total 9 parts of the triangle

⅓ of 9 is 3.

Therefore shade any 3 parts.

13.2

Step-by-step explanation:

r = 2.1

circumference of circle = 2πr

=2 × 22/7 × 2.1

= 13.2

Answer:

y = 4x - 3

Step-by-step explanation:

(-2, -11) & (3, 9)

First you want to find the slope of the line that passes through these points. To find the slope of the line, we use the slope formula: (y₂ - y₁) / (x₂ - x₁)

Plug in these values:

(9 - (-11)) / (3 - (-2))

Simplify the parentheses.

= (20) / (5)

Simplify the fraction.

20/5

= 4

This is your slope. Plug this value into the standard slope-intercept equation of y = mx + b.

y = 4x + b

To find b, we want to plug in a value that we know is on this line: in this case, I will use the second point (3, 9). Plug in the x and y values into the x and y of the standard equation.

9 = 4(3) + b

To find b, multiply the slope and the input of x(3)

9 = 12 + b

Now, subtract 12 from both sides to isolate b.

-3 = b

Plug this into your standard equation.

y = 4x - 3

This is your equation.

Check this by plugging in the other point you have not checked yet (-2, -11).

y = 4x - 3

-11 = 4(-2) -3

-11 = -8 - 3

-11 = -11

Your equation is correct.

Hope this helps!

Answer:

c) k = - 2

Step-by-step explanation:

Equação de Segundo Grau:

Uma equação de segundo grau tem o seguinte formato:

[tex]ax^2 + bx + c = 0[/tex]

Em que [tex]a \neq 0[/tex]

Caso tenham potências de x maiores que dois, o termo que multiplica essas potências deve ser 0.

Neste exercício:

Tem-se a seguinte equação:

[tex](k^2 - 4)x^3 + (k-2)x^2 + 7x - 8 = 0[/tex]

Para ser do segundo grau, necessita-se que:

[tex]k^2 - 4 = 0[/tex]

[tex]k^2 = 4[/tex]

[tex]k = \pm \sqrt{4}[/tex]

[tex]k = \pm 2[/tex]

E também

[tex]k - 2 \neq 0[/tex]

[tex]k \neq 2[/tex]

A intersecção entre [tex]k = \pm 2[/tex] e [tex]k \neq 2[/tex] é [tex]k = -2[/tex], e portanto a resposta é dada pela alternativa c.

O valor de k que torna a equação uma equação de segundo grau para a equação considerada é dado por: Opção C: k = -2

For a single variable equation, degree of an equation is the highest power the variable possess, such that its coefficient is non-zero.

A second degree equation is that equation where the maximum power involved in that equation is 2 (such that its coefficient is non-zero).

For this case, the considered equation is:

(k² – 4) x³ + ( k – 2 )x² + 7x - 8 = 0

For making it second degree, we would have to vanish the coefficient of x³, and by vanish, we mean we would like to make it 0, but simultaneously, we'd have to take care that coefficient of x² doesn't become 0 as we need it to be non-zero.

So we get two mathematical statements:

or

[tex](k^2 - 4) = 0\\k - 2 \neq 0[/tex]

Taking first equation, we get:

[tex]k^2 -4 = 0\\k^2 = 4\\k = \pm \sqrt{4} = \pm 2[/tex]
Thus, k = 2 or k = -2

From second statement, we get:

[tex]k -2 \neq 0\\\text{adding 2 on both the sides}\\k \neq 2[/tex]

Thus, from both the results, we conclude that k = -2 is correct choice.

Thus, the value of k that makes the equation a second degree equation for the considered equation is given by: Option C: k = -2

Learn more about second degree equation here:
https://brainly.com/question/17167165

It will take 36 days for 20 men to complete the building

if you times the two questions the answer is 1.5700 I hope this helps because I'm not that good at math! :)

Answer:

C. Disagree

Explanation:

Because when you're adding mixed fractions, you don't just make them/find there denominator's equivalent/LCM and them, you have to first convert them to improper fractions.

Hope this helps