I just need 6 7 and 8 please

Fairly easy question

nexus9112 [7]

Answer:

C. x=3 and y=−6

Step-by-step explanation:

Step: Solvex−2y=15for x:

x−2y=15

x−2y+2y=15+2y(Add 2y to both sides)

x=2y+15

Step: Substitute2y+15forxin2x+4y=−18:

2x+4y=−18

2(2y+15)+4y=−18

8y+30=−18(Simplify both sides of the equation)

8y+30+−30=−18+−30(Add -30 to both sides)

8y=−48

8y8=−488(Divide both sides by 8)

y=−6

Step: Substitute−6foryinx=2y+15:

x=2y+15

x=(2)(−6)+15

x=3(Simplify both sides of the equation)

Answer:

x=3 and y=−6

7 0

3 years ago

Which expression is a sum of cubes?

luda_lava [24]

$-27x^9+y^6 = (-3)^3\times \Big(x^3\Big)^3 + \Big(y^2\Big)^3=\boxed{\Big(-3x^3\Big)^3+\Big(y^2\Big)^3}$

5 0

3 years ago

Read 2 more answers

Simplify The Following: -4h + 22 + 6h -7

Vikentia [17]

Answer:

2h + 15

Step-by-step explanation:

-4h + 22 + 6h -7

2h + 22 - 7

2h + 15

3 0

3 years ago

A trapezoid is shown. The lengths of the bases are 4 and 8. The height of the altitude is 4. What is the area of the trapezoid?

katrin [286]

It would be 24 units^2 because the middle is 16 and the two triangles are 4 each

3 0

3 years ago

Read 2 more answers

Find the vertex of the graph of the function.

lapo4ka [179]

Answer: The correct options are (1) (5,10), (2) (3,-3), (3) x = -1, (4) $y=(x+2)^2+3$ , (5) 21s and (6) 0, -1, and 5.

Explanation:

Te standard form of the parabola is,

$f(x)=a(x-h)^2+k$ .....(1)

Where, (h,k) is the vertex of the parabola.

(1)

The given equation is,

$f(x)=(x-5)^2+10$

Comparing this equation with equation (1),we get,

$h=5$ and $k=10$

Therefore, the vertex of the graph is (5,10) and the fourth option is correct.

(2)

The given equation is,

$f(x)=3x^2-18x+24$

$f(x)=3(x^2-6x)+24$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $9$ . Since there is factor 3 outside the parentheses, so subtract three times of 9.

$f(x)=3(x^2-6x+9)+24-3\times 9$

$f(x)=3(x-3)^2-3$

Comparing this equation with equation (1),we get,

$h=3$ and $k=-3$

Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.

(3)

The given equation is

$f(x)=4x^2+8x+7$

$f(x)=4(x^2+2x)+7$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $1$ . Since there is factor 4 outside the parentheses, so subtract three times of 1.

$f(x)=4(x^2+2x+1)+7-4$

$f(x)=4(x+1)^2+3$

Comparing this equation with equation (1),we get,

$h=-1$ and $k=3$

The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.

(4)

The given equation is,

$y=x^2+4x+7$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $2^2$ .

$f(x)=x^2+4x+4+7-4$

$f(x)=(x+2)^2+3$

Therefore, the correct option is 4.

(5)

The given equation is,

$h=-16t^2+672t$

It can be written as,

$h=-16(t^2-42t)$

It is a downward parabola. so the maximum height of the function is on its vertex.

The x coordinate of the vertex is,

$x=\frac{b}{2a}$

$x=\frac{42}{2}$

$x=21$

Therefore, after 21 seconds the projectile reach its maximum height and the correct option is first.

(6)

The given equation is,

$f(x)=3x^3-12x^2-15x$

$f(x)=3x(x^2-4x-5)$

Use factoring method to find the factors of the equation.

$f(x)=3x(x^2-5x+x-5)$

$f(x)=3x(x(x-5)+1(x-5))$

$f(x)=3x(x-5)(x+1)$

Equate each factor equal to 0.

$x=0,-1,5$

Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.

3 0

3 years ago