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2 years ago

11

Students collected 600 cans for the canned food drive.

1 answer:

BartSMP [9]2 years ago

5 0

Answer:

they'll need 200 more cans

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What is (f ° g)(2) and (g ° f)(2) with the set of points f: {(0,4), (1,2), (2,0), (3,2), (4,4)} and g: {(0,4), (1,3), (2,2), (3,

kumpel [21]

Answer:

$\large\boxed{(f\circ g)(2)=0}\\\boxed{(g\circ f)(2)=4}$

Step-by-step explanation:

$f:\{(0,4), (1,2), \underline{(2,0)}, (3,2), (4,4)\}\\\\g:\{\underline{(0,4)}, (1,3), \underline{(2,2)}, (3,1), (4,0)\}\\\\(f\circ g)(x)=f\bigg(g(x)\bigg)\\\\g(2)=2\qquad/from\ (2,2)/\\f\bigg(g(2)\bigg)=f(2)=0\qquad/from\ (2,0)/\\\\(g\circ f)(x)=g\bigg(f(x)\bigg)\\\\f(2)=0\qquad/from\ (2,0)/\\g\bigg(f(2)\bigg)=g(0)=4\qquad/from\ (0,4)/$

7 0

2 years ago

What is the ricipical of -2/9

Solnce55 [7]

The recipricol of -2/9 is 9/2, or 4 1/2

remember to change the sign (forgot to myself)

hope this helps

7 0

3 years ago

Read 2 more answers

Given right triangle JKM, which correctly describes the locations of the sides in relation to ∠J?

dangina [55]

It would be opposite from side KM since the side correlated to the angle across it

7 0

3 years ago

Read 2 more answers

Write the equation of the line that passes through (-1, 3) and is parallel to the line y = -x + 1.

ra1l [238]

Answer:

Step-by-step explanation:

y = -x + 1......same as y = -1x + 1.....in y = mx + b form, ur slope is in the m position....so the slope of this line is -1.

a parallel line will have the same slope.

(-1,3)...x = -1 and y = 3

slope(m) = -1

y = mx + b.....we have x,y, and m....now we need to find y int (b)

3 = -1(-1) + b

3 = 1 + b

3 - 1 = b

2 = b

so ur parallel equation is : y = -1x + 2...or just y = -x + 2

7 0

2 years ago

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

Aliun [14]

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

<u>Solution:</u>

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;

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

<em><u>Therefore, the two roots are:</u></em>

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

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