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3 years ago

6

200 grams of fertilizer are used for each tree in an orchard containing 1200 trees. at $2.75 per kilogram of fertilizer, how muc

h does it cost to fertilize the orchard ?

1 answer:

stiks02 [169]3 years ago

4 0

To determine the cost to fertilize the trees, you need to figure out how many kilograms of fertilizer are needed. With the information given, you can determine The number of grams needed to fertilize 1200 trees. To do this you would multiply 200 times 1200. This equals 240,000 grams. To convert this to kilograms, divide 240000g by 1000 g. Every group of 1000 g is 1 kg. The answer is 240 kilograms. Multiply 240 kg by the price of $2.75 per kilogram to get $660 as the cost.

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Graph: - 4x + 4y > 20 & 4x + 4y < -12<br>please help !!!!!<br>

VARVARA [1.3K]

I’m not totally sure about the less and greater signs because I’m not sure you can graph using those signs but either way their are your line equations

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3 years ago

Help me Please!! Ive got a deadline!! Thanks! (:

saul85 [17]

The equation in standard form is 2x^2 + 7x - 15=0. Factoring it gives you (2x-3)(x+5)= 0. That's the first one. The second one requires you to now your formula for the axis of symmetry which is x = -b/2a with a and b coming from your quadratic. Your a is -1 and your b is -2, so your axis of symmetry is
x= -(-2)/2(-1) which is x = 2/-2 which is x = -1. That -1 is the x coordinate of the vertex. You could plug that back into the equation and solve it for y, which is the easier way, or you could complete the square on the quadratic...let's plug in x to find y. -(-1)^2 - 2(-1)-1 = 0. So the vertex is (-1, 0). That's the first choice given. For the last one, since it is a negative quadratic it will be a mountain instead of a cup, meaning it doesn't open upwards, it opens downwards. Those quadratics will ALWAYS have a max value as opposed to a min value which occurs with an upwards opening parabola. This one is also the first choice because of the way the equation is written. There is no side to side movement (the lack of parenthesis tells us that) so the x coordinate for the vertex is 0. The -1 tells us that it has moved down from the origin 1 unit; hence the y coordinate is -1. The vertex is a max at (0, -1)

3 0

3 years ago

6x+ 6x=<br> I need the answer pleasee

kipiarov [429]

Answer:

The answer is 12x because 6x and 6x are like terms (same variable)

5 0

3 years ago

Read 2 more answers

Find solution to logarithm.

sveta [45]

$\log5x+\log(x-1)=2\\
D:5x>0 \wedge x-1>0\\
D:x>0 \wedge x>1\\
D:x>1\\
\log5x(x-1)=2\\
10^2=5x^2-5x\\
5x^2-5x-100=0\\
x^2-x-20=0\\
x^2-5x+4x-20=0\\
x(x-5)+4(x-5)=0\\
(x+4)(x-5)=0\\
x=-4 \vee x=5\\
-4\not \in D\Rightarrow x=5
$

8 0

3 years ago

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Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

5 0

3 years ago

Read 2 more answers