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3 years ago

Help please i need it done quickly

Mathematics

2 answers:

Mariulka [41]3 years ago

8 0

5.24 and 2.85 I think that’s correct

Katena32 [7]3 years ago

4 0

First person who responded is right

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One thousand and twenty students participated in the school math contest.

Tema [17]

Answer:

sample 1: 75

sample 2: 86

Step-by-step explanation:

6 0

2 years ago

The conversion rate for US Dollars to Costa Rican Colones is 1 USD = 518 Colones.

allsm [11]

The cost of 0.5 kg of bananas is 393.60 Colones as per the given conversion rates

Conversion rate of 1 USD to Costa Rican Colones = 518 Colones

The conversion rate of kg to pounds given in the question: 1 kg = 2.2025 lbs

Cost of one pound of bananas = $0.69

Bananas required to be purchased = 0.5kg

Converting 0.5kg bananas to pounds = 0.5*2.2025 = 1.10125 pounds

Cost of 1.10125 pound of bananas in dollars = 1.10125*0.69 = 0.7598

Cost of 1.1025 pounds of bananas in Colones = 0.7598*518 = 393.60 Colones

Hence, the cost is 393.60

Therefore, the cost of 0.5 kg bananas in Colones is 393.60 Colones

Learn more about conversion rate:

brainly.com/question/2274822

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5 0

1 year ago

Is the result of Double 5 and add 1 the same as the result of Add 1 to 5; then double the result?

just olya [345]

B. Double 5 plus 1 equals 11 and add 1 to 5 doubled equals 12.

7 0

3 years ago

Read 2 more answers

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

The area of a rectangle is 4.9 square units. The length is 2.5 units and the width is y units. What is the value of y?

iris [78.8K]

1.96, if you multiply that by the length, (which is 2.5) then you will get the area of 4.9 hope this helps

4 0

3 years ago

Read 2 more answers