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2 years ago

A. How are the functions y = x and y = x + 4 related? How are their graphs related?

Mathematics

1 answer:

kicyunya [14]2 years ago

5 0

Step-by-step explanation:

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You are to construct an open rectangular box with a square base and a volume of 48 ft^3. If material for the bottom costs $6/ft^

Paha777 [63]

I think it is 5/12 I don’t know that

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3 years ago

What is the solution i need it fast!!!

zloy xaker [14]

Answer:

x < 3

Step-by-step explanation:

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3 years ago

I need help with this geometry question.

sashaice [31]

Answer:

Step-by-step explanation:

a). There are 8 points in the figure attached.

b). There are 9 lines in the given figure.

c). There are 5 planes in the figure attached.

d). Three collinear points are D,G and F.

e). Four co-planar points are G, F, H and C.

f). Intersection of planes ABC and ABE is the common line AB.

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3 years ago

Brandon buys a radio for $43.75 in a state where the sales tax is 7%.

Nastasia [14]

Answer:

Tax: $3.06

Step-by-step explanation:

7% of 43.75 = 0.07 × 43.75 = 3.0625

43.75 increase 7% =

43.75 × (1 + 7%) = 43.75 × (1 + 0.07) = 46.8125

Radio: $43.75

Tax: $3.06

Total: $46.81

4 0

2 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

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

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago