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3 years ago

Karin wants to use the distributive property to mentally find the value of 19 x 42+ 19x 58. Which expression can she use?

Mathematics

2 answers:

olasank [31]3 years ago

7 0

Answer:

19(42 + 58)

Step-by-step explanation:

Assuming x = multiply, you are trying to solve the distributive property. Divide common factors from both terms (in this case 19).

kifflom [539]3 years ago

4 0

Answer:

19(42 + 58)

Step-by-step explanation:

19 * 42 + 19 * 58 =

= 19 * (42 + 58) <------- this is the expression

= 19 * 100

= 1900

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Write a whole number and a fraction greater than one to name the part filled.

san4es73 [151]

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3 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

18. Jackie drives 160 miles from Gary, Indiana to

astra-53 [7]

Answer:

11 hours and 15 minutes (11.25)

Step-by-step explanation:

make a ratio

160:4.5

240/160=1.5

so 1 and 1/2 intervals of 4.5 hours

4.5+4.5= 9

4.5/2=2.25

9+2.25=11.25

6 0

3 years ago

If you answer both I will give Brainliest they have to be correct.Good luck

svetlana [45]

Answer:

Step-by-step explanation:

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3 years ago

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William has been contracted to paint a school classroom. The classroom is 20 m long, 15 m wide and 5 m high. There are four wind

valina [46]

Answer:

Step-by-step explanation:

l -> length

b -> width

h -> height

Find the area of four walls and ceiling. then subtract the area of four windows and a door form that area.

Area of four walls + <u>ceiling</u> = 2( lh + bh) +<u>lb</u>

= 2*(20*5 + 15*5) + 20*15

= 2( 100 + 75) + 300

= 2* 175 + 300

= 350 +300

= 650 sq m

Area of window = 2 *3 = 6 sq.m

Area of four windows = 4*6 = 24 sq.m

Area of door = 2 * 1 = 2 sq.m

Area of four walls excluding 4 windows and door = 650 - 24 - 2 = 624 sq.m

Cost of painting = 624 * 6.50

= $ 4056

3 0

3 years ago

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