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3 years ago

Write an expression to represent: One more than the product of one and a number x.

Mathematics

2 answers:

stepan [7]3 years ago

6 0

The answer to this question is 2x due to 2 being one more than 1 which is x

inessss [21]3 years ago

6 0

1x=x(1) because you don't actually solve it. and if it's one more then a product then its one times a product.

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Volume is the total number of cubic units contained inside a three-dimensional figure.

Tasya [4]

What I think is true!!

7 0

3 years ago

A company has $6582 to give out in bonuses. An amount is to be given out equally to each of the 32 employees.

expeople1 [14]

A) His estimate is incorrect, since he has increased a figure to the amount that each employee would receive.

B) Rounded to the nearest dollar, each employee would receive $ 209.

Since a company has $ 6582 to give out in bonuses, and an amount is to be given out equally to each of the 32 employees, and A) a manager, Jake reasoned that since 32 goes into 64 twice, each employee will get about $ 2000 , to determine if his estimate is correct, and B) determine how much will each employee receive, rounded to the nearest whole dollar, the following calculations must be performed:

2000 x 32 = 64000
200 x 32 = 6400

Therefore, his estimate is incorrect, since he has increased a figure to the amount that each employee would receive.

6582/32 = X
208.68 = X

Therefore, rounded to the nearest dollar, each employee would receive $ 209.

Learn more about maths in brainly.com/question/8865479

4 0

2 years ago

Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an

Natasha_Volkova [10]

Answer:

The volume of the solid is 243 $\sqrt{2} \ \pi$

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0 ≤ ρ² ≤ 81

0 ≤ ρ ≤ 9

The intersection that takes place in the sphere and the cone is:

$x^2 +y^2 ( \sqrt{x^2 +y^2 })^2 = 81$

$2(x^2 + y^2) =81$

$x^2 +y^2 = \dfrac{81}{2}$

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

$z = \sqrt{x^2+y^2}$

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; $\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}$

Thus, volume: $V = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4} \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho ^2 \ sin \phi \ d\rho \ d \theta \ d \phi$