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3 years ago

What is the step by step equation for −4+12=−4? thanks!

Mathematics

2 answers:

m_a_m_a [10]3 years ago

6 0

Answer:

the actually answer is -8

Step-by-step explanation:

GuDViN [60]3 years ago

3 0

No solution to solve this answer it’s false

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roger bought 8 small bricks for his garden for $5.46. the total weight of the bricks was 6 1/4 pounds how many pounds does one b

mestny [16]

Answer:

0.78125 pounds

Step-by-step explanation:

the weight of 8 brick is 6.25 pound

divide the weight by 8 to take the weight for the one brick

3 0

3 years ago

Solve the system by substitution...

Nataly_w [17]

Answer: 14

Step-by-step :x +3 y = 15 , 2 x -y =14

but let me know if it's right ok

6 0

2 years ago

Find all solutions in the interval [0,2pi)<br> sec^2x + 2tanx = 3

Marysya12 [62]

Tan^2x + 1 + tanx - 3 = 0 tan^2x + tanx - 2 = 0 (tanx + 2)(tanx - 1) = 0 tanx = -2 or tanx = 1 x = 2.034 radians or x = 5.176 radians or x = pi/4 radians or x = 5pi/4 radians

6 0

3 years ago

The volume of a sphere whose diameter is 18 centimeters is _ cubic centimeters. If it’s diameter we’re reduced by half, it’s vol

kaheart [24]

<h2>Answer:</h2>

<u>First Part</u>

Given that

$Volume = \frac{4}{3} \pi r^{3}$

We have that

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{Diameter}{2})^{3} = \frac{4}{3} \pi 9^{3} = 972\pi cm^{3} \approx 3053.63 cm^{3}$

<u>Second Part</u>

Given that

$Volume = \frac{4}{3} \pi r^{3}$

If the Diameter were reduced by half we have that

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{r}{2}) ^{3} = \frac{\frac{4}{3} \pi r^{3}}{8}$

This shows that the volume would be $\frac{1}{8}$ of its original volume

<h2>Step-by-step explanation:</h2>

<u>First Part</u>

Gather Information

$Diameter = 18cm$

$Volume = \frac{4}{3} \pi r^{3}$

Calculate Radius from Diameter

$Radius = \frac{Diameter}{2} = \frac{18}{2} = 9$

Use the Radius on the Volume formula

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi 9^{3}$

Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s

$Volume = \frac{4}{3} \pi 9^{3} = \frac{4}{3} \pi 9 * 9 * 9 = \frac{4}{3} \pi 9 * 9 * 3 * 3$

We can see now that one of the 3s can be already divided by the 3 in the denominator

$Volume = \frac{4}{3} \pi 9 * 9 * 3 * 3 = 4 \pi 9 * 9 * 3$

Finally, since we can't simplify anymore we just calculate it's volume

$Volume = 4 \pi 9 * 9 * 3 = 12 \pi * 9 * 9 = 12 * 81 \pi = 972 \pi cm^{3}$

$Volume \approx 3053.63 cm^{3}$

<u>Second Part</u>

Understanding how the Diameter reduced by half would change the Radius

$Radius =\frac{Diameter}{2}\\\\If \\\\Diameter = \frac{Diameter}{2}\\\\Then\\\\Radius = \frac{\frac{Diameter}{2} }{2} = \frac{\frac{Diameter}{2}}{\frac{2}{1}} = \frac{Diameter}{2} * \frac{1}{2} = \frac{Diameter}{4}$

Understanding how the Radius now changes the Volume

$Volume = \frac{4}{3}\pi r^{3}$

With the original Diameter, we have that

$Volume = \frac{4}{3}\pi (\frac{Diameter}{2}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{2^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{2 * 2 * 2} = \frac{4}{3}\pi \frac{Diameter^{3}}{8}\\\\$

If the Diameter were reduced by half, we have that

$Volume = \frac{4}{3}\pi (\frac{Diameter}{4}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{4^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 4 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 2 * 2 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{8 * 8} = \frac{\frac{4}{3}\pi\frac{Diameter^{3}}{8}}{8}$

But we can see that the numerator is exactly the original Volume!

This shows us that the Volume would be $\frac{1}{8}$ of the original Volume if the Diameter were reduced by half.

3 0

2 years ago

These roots of the polynomial equation x^4-4x^3-2x^2+12x+9=0 are 3,-1,-1. Explain why the fourth root must be a real number. Fin

Alex787 [66]

Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.

Because we know 3 and -1 (multiplicity 2) are both roots, the last root $r$ is such that we can write

$x^4-4x^3-2x^2+12x+9=(x-3)(x+1)^2(x-r)$

There are a few ways we can go about finding $r$ , but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be $(-3)(1)(1)(-r)=3r$ .

Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

$3r=9\implies r=3$

so the missing root is 3.

Other things we could have tried that spring to mind:

- three rounds of division, dividing the quartic polynomial by $(x-3)$ , then by $(x+1)$ twice, and noting that the remainder upon each division should be 0

- rational root theorem

4 0

3 years ago

Read 2 more answers