A line y=mx+b passes through point (−6, 4) and is perpendicular to y=−12x+10. What is the value of b?

The sum of three consecutive even integers is 258. write an equation ad solve to find the integers

slavikrds [6]

The consecutive integers are 85,86,87

Explanation:

n

:

the first number

n

+

1

:

the second number

n

+

2

:

the third number

n

+

(

n

+

1

)

+

(

n

+

2

)

=

258

3

n

+

3

=

258

3

n

=

258

−

3

n

=

255

n

=

255

3

n

=

85

n

+

1

=

85

+

1

=

86

n

+

2

=

85

+

2

=

87

8 0

3 years ago

Read 2 more answers

Use the function f(x) = x2 - 6x + 3 and the graph of g(x) to determine the difference between the maximum value of g(x) and the

motikmotik

Answer: Use the function f(x)=x2-2x+8 and the graph of g(x) to determine the difference betw een the maximum value of g(x) and the minimum value of f(x).

5 0

4 years ago

X-5=5x-13<br> i need the steps

posledela

Answer:

x=2

Step-by-step explanation:

x-5=5x-13

5x-x=-5+13

4x=8

x=8/4

x=2

4 0

3 years ago

Read 2 more answers

The numerator of a fraction is 15 less than twice its denominator. If the numerator is increased by 5 and the denominator is inc

denis-greek [22]

Answer:

7/11

Step-by-step explanation:

Let d represent the original denominator. Then the original numerator is ...

2d-15

The new numerator is ...

(2d-15) +5

and the new denominator is ...

d+7

The ratio of these is 2/3, so we have ...

$\dfrac{2d-15+5}{d+7}=\dfrac{2}{3}\\\\3(2d-10)=2(d+7) \quad\text{cross multiply}\\\\4d=44 \quad\text{add 30-2d}\\\\d=11\\\\2d-15=2(11)-15=7$

The original fraction is 7/11.

7 0

3 years ago

What numbers are zeros of g(x) = x^2 - 2x - 4? take your time if you need to!

ipn [44]

Answer:

$x = 1 + \sqrt5, x = 1 - \sqrt5$

Step-by-step explanation:

Hello!

We can solve the quadratic by using the quadratic formula.

Standard form of a quadratic: $ax^2 + bx + c = 0$

Quadratic Formula: $x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

Given our Equation: $g(x) = x^2 - 2x - 4$

a = 1
b = -2
c = -4

Plug the values into the equation and solve.

<h3>Solve</h3>

$x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-(-2)\pm\sqrt{(-2)^2 - 4(1)(-4)}}{2(1)}$
$x = \frac{2\pm\sqrt{4 +16}}{2}$
$x = \frac{2\pm\sqrt{20}}{2}$
$x = \frac{2\pm\sqrt{4 * 5}}{2}$
$x = \frac{2\pm(\sqrt4 * \sqrt5)}{2}$
$x = \frac{2\pm2\sqrt5}{2}$
$x = 1 + \sqrt5, x = 1 - \sqrt5$

7 0

2 years ago