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3 years ago

Which is the graph of the system of equation y=3x+6 and y=3x+1?

Mathematics

2 answers:

Fynjy0 [20]3 years ago

7 0

Answer:

the answer is C

Pavlova-9 [17]3 years ago

5 0

Answer:

straight line with y intercept of 6 and positive slope of 3

straight line with y intercept of 1 and positive slope of 3

Step-by-step explanation:

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Adult tickets to a play cost $24. Tickets for children cost $18. Tickets for a group of 13 people cost $282. Write and solve a s

Elan Coil [88]

Let number of Adult ticket be x
and number of Child ticket be y
so
$282 = $24x + $18y.
:)

4 0

3 years ago

Read 2 more answers

The prism shown has a volume of 35cm³. Work out the height of the triangular cross section. The length going back is 7cm and the

Gre4nikov [31]

Answer:

2.5 cm

Step-by-step explanation:

The volume of a triangular prism is cross section × length.

The cross section area is the triangle area. The triangle length is 4 cm, the height is not given. The length of the triangular prism is 7cm.

4 × h × 1/2 × 7 = 35

14h = 35

h = 35/14

h = 5/2

The height of the triangular cross section is 2.5 cm.

3 0

4 years ago

What is the nth term of the geometric sequence 4, 8, 16, 32, ... ?

Morgarella [4.7K]

Geometric is
$a_n=a_1r^{n-1}$
an=nth term
a1=first term
r=common ratio
n=which term

commmon ratio is like
a1 times r=a2
4 to 8
4 times 2=8
r=2
first term is 4
therefor

$a_n=4(2)^{n-1}$
the nth term is $4(2)^{n-1}$

3 0

4 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

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

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$