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3 years ago

The expanded fom of (a-b)²

Mathematics

2 answers:

Alexxandr [17]3 years ago

6 0

-2ab-b^
Hope it helps ur question

emmainna [20.7K]3 years ago

4 0

Answer:

Step-by-step explanation:

$(a-b)^{2}$ = $a^{2}$ - 2ab + $b^{2}$

Hope this helps

plz mark as brainliest!!!!!

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How to do an angle of impact

oksian1 [2.3K]

Answer:

Measure the length and width of the splatter. Divide the width of the splatter by its length.

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3 years ago

Part of your job as manager is to count the cash for the daily bank deposit. When you count the cash, you have 47 ones, 22 fives

gavmur [86]

The total amount of the deposit is $592.92.

<h3>What is the total amount of the deposit?</h3>

The total amount of the deposit can be determined by adding all the money deposited together.

(47 x 1) + (22 x 5) + (9 x 10) + (17 x 20) + (67 x 0.01) + (0.05 x 12) + (9 x 0.1) + (15 x 0.25)

47 + 110 + 90 + 340 + 0.67 + 0.60 + 0.9 + 3.75 = $592.92

To learn more about addition, please check: brainly.com/question/19628082

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2 years ago

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

A patio 2 pairs of two parallel sides and 2 pairs of sides that are the same length there are 4 square corners what shape is the

Tems11 [23]

Answer:

The patio is a rectangle

Step-by-step explanation:

Since it is given that we have 2 pairs of two parallel sides, which implies 4 sides, this thus shows that the shape is a quadrilateral since, only quadrilaterals have 4 sides. Also, 2 pairs of sides are the same length, and are parallel, it shows that the shape is likely a rectangle.

Finally, there are 4 square corners on the shape which implies 4 right angles. This confirms that our shape is a rectangle since only a rectangle has 2 pairs of two parallel sides and 2 pairs of sides that are the same length there are 4 square corners.

So, the patio is a rectangle.

3 0

3 years ago

3×-7 equivalent expressions A. -21 B. 4 C. -4 D. 21

USPshnik [31]

Answer:

-21

Step-by-step explanation:

3x7=21 and a negative x a positive is always negative

6 0

2 years ago

The expanded fom of (a-b)²​

The expanded fom of (a-b)²