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2 years ago

HELP PLEASE WILL GIVE BRAINLIST, I DONT HAVE MUCH TIME.IF I PASS THIS I GET TO GO TO DISNEY

Mathematics

1 answer:

Verdich [7]2 years ago

6 0

Answer: 41

Step-by-step explanation:

I think it’s forty one because each child painted one and there are forty one flower pots. Besides that much information wasn’t given so it is most likely 41 sorry if it wasn’t much help

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How do you do 39 divided by 936 long division

viva [34]

936 39 = 24

24 = 240 to the nearest tenth

24 = 24 to the nearest hundredth

24 = 24 to the nearest thousandth

= 0 to the nearest tenth

= 0 to the nearest hundredth

= 0 to the nearest thousandth

8 0

2 years ago

Jason and Jared have a lawn-mowing service. They charge $20 per hour. One lawn was 150 ft. by 100 ft. and it took 11⁄2 hours to

rusak2 [61]

Answer:110$

Step-by-step explanation: well 11/2 hours equals 19800 seconds and 19800 seconds equals 5 hours and a half so 20 x 5 equals 100 and ya so and half 20 is 10 so 110$

3 0

2 years ago

Read 2 more answers

Solve for x: −4x+6(6−2x)=−7x−36

ValentinkaMS [17]

Answer:

x = 8

Step-by-step explanation:

-4x + 6(6 - 2x) = -7x - 36

-4x + 36 - 12x = -7x - 36

-16x + 36 = -7x - 36

+36 +36

-16x + 72 = -7x

+16x +16x

72 = 9x

divide by 9

x = 8

4 0

2 years ago

Read 2 more answers

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 length of X and length of Y

luda_lava [24]

Answer:

It is a right angled triangle.

tan32 = p/b

or, tan32 = x/5

or, tan32×5 = x

so, x = 3.124

for 5 no. the answer is a. 3.124

For y,

cos32 = b/h

or, cos32 = 5/y

or, y = 5/cos32

so, y = 5.895

for 6 no. 5.895 is closest to 5.994, so the answer is c. 5.994.

4 0

2 years ago