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2 years ago

Write a subtraction problem using two fractions with unlike denominators that have a difference of 2/5

Mathematics

1 answer:

timurjin [86]2 years ago

6 0

2/5 = 8/20
8/20 + 5/20 = 13/20

13/20 - 1/4 = 2/5

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11 is 10% more than ?

djyliett [7]

Answer: 21

Step-by-step explanation:

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2 years ago

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A store pays $20 for a hat. to make a profit the store prices the hat 60% higher. later a sale reduce the price by 10%. finally,

Maslowich

The clearance price of hat is $17.28

Step-by-step explanation:

Price paid by store = $20

Increment = 60% of price paid

Increment = $\frac{60}{100}*20 = \frac{1200}{100}$

Increment = $12

Increased price = 20+12 = $32

Sale = 10%

Discount amount = 10% of increased price

Discount amount = $\frac{10}{100}*32=\frac{320}{100}$

Discount amount = $3.2

Sale price = 32 - 3.2 = $28.8

Clearance sale = 40%

Discount of clearance sale = 40% of sale price

Discount = $\frac{40}{100}*28.8=\frac{1152}{100}$

Discount = $11.52

Clearance price = sale price - Discount of clearance sale

Clearance price = 28.8 - 11.52 = $17.28

The clearance price of hat is $17.28

Keywords: percentage, subtraction

Learn more about subtraction at:

#LearnwithBrainly

6 0

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

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

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3 years ago

Explain why multiplying two irrational numbers could result in either an irrational number or a rational number.

Nimfa-mama [501]

Because if you multiply 2 different irrational numbers (like the following)
$\sqrt{3} \times \sqrt{17}$
You will get an irrational answer.

BUT
if you multiply the same square root (which are both irrational, you get a rational number)

$\sqrt{3} \times \sqrt{3} = 3$

7 0

3 years ago

20 POINTS FOR THE FIRST PERSON TO ANSWER

MAVERICK [17]

Answer: its A

Step-by-step explanation:

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3 years ago

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