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3 years ago

The label on a \frac{1}{4}

Mathematics

1 answer:

alexgriva [62]3 years ago

4 0

Answer:

$183.63$ square feet area is covered by seeds per pound

Step-by-step explanation:

Given

$41\frac{1}{4}$ pound bag of seed cover an area of $7575$ square feet

$\frac{165}{4}$ pound bag of seed cover an area of $7575$ square feet

$7575$ square feet area is covered by $\frac{165}{4}$ pound bag of seed

Area in square feet covered by one pound bag of seed

$= \frac{7575}{\frac{165}{4} } \\= \frac{7575 * 4}{165} \\= 183.63$ square feet

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Select the equation that contains the point (3, -9), and in which the graph of the line has a positive slope.

Kamila [148]

First thing to do is to solve each of these for y. The first one is y=-4x-3; the second one is y=4x-21; the third one is y=4x+21; the fourth one is y=-4x+3. From that you can tell the positive slopes are found in the second and third equations. Those are the ones we will test now for the point (3, -9). y=-9 and x=3, so let's fill in accordingly. The second equation filled in is -9=4(3)-21. Does the left side equal the right when we do the math? -9=12-21 and -9=-9. So the second one works. Just for the sake of completion, let's do the same with the third: -9=4(3)+21. Does -9=12+21? Of course it doesn't. Our equation is the second one above, y+9=4(x-3).

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3 years ago

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allochka39001 [22]

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3 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Solve the system by substitution. y = -3x +4 y = 2

erma4kov [3.2K]

Answer:

(2/3,2)

Step-by-step explanation:

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3 years ago

Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

Answer:

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

Solution:

Need to solve given equation using quadratic formula.

$-3 x^{2}-x-3=0$

General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

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

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$

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4 years ago