All categories

2 years ago

HELP ASAP I'M FAILING AND I DON'T UNDERSTAND THIS

1 answer:

7 0

The intercept can be found when all other variables are equated to zero.
x-intercept when y = 0 and z = 0: 8x + 6*0 + 3*0 = 24 gives x = 3
y-intercept when x = 0 and z = 0: 8*0 + 6y + 3*0 = 24 gives y = 4
z-intercept when x = 0 and y = 0: 8*0 + 6*0 + 3z = 24 gives z = 8

The intercepts are (3, 0, 0), (0, 4, 0), and (0, 0, 8).

You might be interested in

ALGEBRA QUESTION PLS HELPS

cluponka [151]

The value of x is –7.

Solution:

Given expression:

$$\left(\frac{1}{x+3}+\frac{6}{x^{2}+4 x+3}\right) \cdot \frac{x+3}{x+1}$

Let us factor $x^2+4x+3$ .

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

Substitute this in the fraction.

$$\left(\frac{1}{x+3}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}$

To make the denominator same, multiply and divide the first term by (x +1).

$$\left(\frac{(x+1)}{(x+1)(x+3)}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}$

Denominators are same, you can add the fractions.

$$\left(\frac{x+1+6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}$

$$\frac{x+7}{(x+1)(x+3)} \cdot \frac{x+3}{x+1}$

Cancel the common term in the numerator and denominator.

$$\frac{x+7}{x+1} \cdot \frac{1}{x+1}$

Multiply the fractions.

$$\frac{x+7}{(x+1)^2}$

$$\frac{x+7}{x^2+2x+1}$

The expression is simplified to one rational expression.

Suppose the expression is equal to 0.

$$\frac{x+7}{x^2+2x+1}=0$

Do cross multiplication.

$${x+7}=0\times (}{x^2+2x+1})$

Any number or variable multiplied by 0 gives 0.

$${x+7}=0$

Subtract 7 from both sides of the equation.

$${x+7-7}=0-7$

x = –7

The value of x is –7.

7 0

2 years ago

I need help with this qustion ?

emmainna [20.7K]

The answer should be 32.

4 0

2 years ago

On the graph shown, what is f(-1)? (5 points) The first segment joints the point negative three comma five exclusively and negat

tatiyna

ANSWER

$f( - 1) = 2$

EXPLANATION

The given function is a piecewise defined function.

$f( - 1)$
is the y-value that corresponds to
$x = - 1.$

To determine that y-value, we need to trace from
$x = - 1$
to where it meets the graph.

We then trace to the y-axis to determine the corresponding y-value as shown in the diagram.

From the graph, we can see that this x-value corresponds to a y-value of
$2$

This implies that, when
$x = - 1,y = 2$

Therefore
$f( - 1) = 2$
See graph.