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2 years ago

Need help ASAP! Please

Mathematics

2 answers:

iren [92.7K]2 years ago

6 0

Answer:

112

Step-by-step explanation:

20 x 20 is 40

6 x 9 is 54

6 x 3 is 18

54 + 18 is 72

72 + 40 is 112

dedylja [7]2 years ago

5 0

Answer:

472

Step-by-step explanation:

Use the order of operations PEMDAS

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

9 + 3 = 12

20^2 = 400

6 * 12 = 72

400 + 72 = 472

hope this helped! :)

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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

Suppose that in a large metropolitan area, 84% of all households have cable tv. Suppose you are interested in selecting a group

Firdavs [7]

Answer:

0.05606

Step-by-step explanation:

This is a binomial probability distribution problem.

84% of all households have cable TV, thus;

p = 0.84

There are six households from this area, thus; n = 6

Formula for this binomial distribution is;

P(X) = C(n, x) × p^(x) × (1 - p)^(n - x)

We want to find the proportion of groups where at most three of the households have cable TV.

Thus, it's is;

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = 0) = C(6, 0) × 0.84^(0) × (1 - 0.84)^(6 - 0) = 0.00001678

P(X = 1) = C(6, 1) × 0.84^(1) × (1 - 0.84)^(6 - 1) = 0.000528

P(X = 2) = C(6, 2) × 0.84^(2) × (1 - 0.84)^(6 - 2) = 0.006963

P(X = 3) = C(6, 3) × 0.84^(3) × (1 - 0.84)^(6 - 3) = 0.04855

Thus;

P(X ≤ 3) = 0.00001678 + 0.000528 + 0.006963 + 0.04855 = 0.05606

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Answer:

Step-by-step explanation:

2x - y when x = 0 and y = -5

2(0) - (-5) =

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