The system of linear equations and the solution for the system is the value. It creates all of the equations true. 2 variables or 2 equations this is the point where the two graphs intersect.

Step-by-step explanation:

Its kina complicated...if you still don't understand don't worry i don't either:/

7 0

3 years ago

Please I need help I don't understand

mario62 [17]

Answer:

Step-by-step explanation:

from the graph of f(x)

when x=1,f(x)=0

or f(1)=0

when f(x)=2,x=2

for g(x)

when x=6,g(x)=16

or g(6)=16

when g(x)=18,x=32

for h(x)

when x=14

h(x)=27x-7

h(14)=27×14-7=7(27×2-1)=7(54-1)=7×53=371

h(x)=-493

27x-7=-493

27 x=-493+7=-486

3 x=-54

x=-18

for p(t)

when t=94

p(t)=24

p(94)=24

p(t)=67

t=31

8 0

3 years ago

NEED THE ANSWER ASAP PLEASE

Elan Coil [88]

Answer:

y=1/2x-3

gradient = 1/2

also should satisfy the given coordinates

6 0

3 years ago

Read 2 more answers

Identify the zeros of the function f(x) = 4x2 − 8x − 1 using the Quadratic Formula. HELP ASAP!!

forsale [732]

$1+\dfrac{\sqrt{5}}{2},1-\dfrac{\sqrt{5}}{2}$

Step-by-step explanation:

The given equation is $4x^{2}-8x-1$

Let $a$ be the coefficient of $x^{2}$

Let $b$ be the coefficient of $x$

Let $c$ be the constant.

Then the roots α,β for the equation $ax^{2}+bx+c$ are $\dfrac{-b+\sqrt{b^{2}-4ac} }{2a},\dfrac{-b-\sqrt{b^{2}-4ac} }{2a}$

So,α= $\frac{-b+\sqrt{b^{2}-4ac} }{2a}=\frac{8+\sqrt{64+16} }{8}=\frac{8+4\sqrt{5}}{8}=1+\frac{\sqrt{5}}{2}$

β= $\frac{-b-\sqrt{b^{2}-4ac} }{2a}=\frac{8-\sqrt{64+16} }{8}=\frac{8-4\sqrt{5}}{8}=1-\frac{\sqrt{5}}{2}$ .

So the roots are $1+\frac{\sqrt{5}}{2},1+\frac{\sqrt{5}}{2}$

5 0

3 years ago