For this exponential function what is the output value (y) when the input value (x) is 0?

three snack bars contain 1/5,0.22,and 19% of their calories from fat. which snack bar contains the least amount of calories from

podryga [215]

19% becuze .22 is 22% and 1/5 is 20%

7 1

3 years ago

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Can someone help? Would be appreciated.

pickupchik [31]

Answer:

WOah

Step-by-step explanation:

5 0

3 years ago

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Having some trouble with this...pls help (in fraction form)

Yuri [45]

Answer:

$\frac{17}{6}$ or $2\frac{5}{6}$

Step-by-step explanation:

1.Find the Least Common Denominator (LCD)

LCD = 6

2.Make the denominators the same as the LCD.

$2+\frac{1*3}{2*3} + \frac{1*2}{3*2}$

3.Simplify. Denominators are now the same

$2+\frac{3}{6} + \frac{2}{6}$

4. Join the denominators

$2+\frac{3+2}{6}$

5.Simplify

$2\frac{5}{6}$ = $\frac{17}{6}$

4 0

3 years ago

Find the roots of:

Elena-2011 [213]

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \ 2x^3-7x^2+8x-3=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

| 2 -7 8 -3

1 | 2 -5 3

| 2 -5 3 0

1 | 2 -3

2 -3 0

So the factorization is (x-1)² (2x-3)=0. So:

$\bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2} }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \ x^3-x^2-4=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

| 1 -1 0 -4

2 | 2 2

1 2 2 0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

$\bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \ 6x^3+7x^2-9x+2=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

| 6 7 9 2

-2 | -12 10 -2

6 -5 1 0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

$\bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}$

$\large\displaystyle\text{$\begin{gathered}\sf \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}$

6 0

2 years ago

What set of reflections would carry rectangle ABCD onto itself?

Dvinal [7]

Answer:

y=x, x-axis, y=x, y-axis

Explanation:

Reflecting the figure across three axes just moves it from one quadrant to another. It does not map the figure to itself.

Reflecting across the line y=x moves it from quadrant II to IV or vice-versa. If it is in quadrant I or III, it stays there. So the sequence of reflections x-axis (moves from I to IV), y=x (moves from IV to II), x-axis (moves from II to III), y=x (stays in III) will not map the figure to itself.

However, the last selection will map the figure to itself. The initial (and final) figure location, and the intermediate reflections are shown in the attached. The figure starts and ends as blue, is reflected across y=x to green, across x-axis to orange, across y=x to red, and finally across y-axis to blue again.

7 0

4 years ago