All categories

1 year ago

PLEASE HELP ME!!!!!!!!!

Mathematics

1 answer:

vitfil [10]1 year ago

8 0

$(x - 4) {}^{2} - 28 = 8 \\ (x - 4) {}^{2} = 8 + 28 \\ (x - 4) {}^{2} = 36$

This must be true for some value of x, since we have a quantity squared yielding a positive number, and since the equation is of second degree,there must exist 2 real roots.

$\sqrt{(x - 4) {}^{2} } = ± \sqrt{36} \\x - 4 = ±6 \\ x _{1}- 4 = 6 \: \: \: \: \: \: \: \: \: \: x _{2}- 4 = - 6 \\ x_{1} = 6 + 4 \: \: \: \: \: \: \: \: \: \: x_{2} = - 6 + 4 \\ x_{1} = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2} = - 2$

Well he started off correct to the point of completing the square.

$(x - 3) {}^{2} = 16 \\ x - 3 = ±4 \\ \: x_{1} - 3 = 4 \: \: \: \: \: \: \: \: \: \: \: \: x_{2} - 3 = - 4 \\ x_{1} = 7 \: \: \: \: \: \: \: \: \: \: x_{2} = - 1$

You might be interested in

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

What is -5/4×1/3 and how do you solve fractions??

kolbaska11 [484]

First off when multiplying fractions you always multiply straight across. so the answer would be -5/12.

6 0

3 years ago

Help me with this problem please.

melomori [17]

Answer:

your the problem

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Which is a negatively skewed distribution?<br> HELP PLEASE

zlopas [31]

Answer:

i think its c

Step-by-step explanation:i think its c because it has dot on the very left

5 0

2 years ago

Plz help !!! timed quiz

Zanzabum

Answer:

6 1/12

Step-by-step explanation:

you would add all the lengths together and then you would have your answer which is 6 1/12

3 0

3 years ago