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3 years ago

Share 270g in the ratio 5 : 4

Mathematics

1 answer:

Phantasy [73]3 years ago

8 0

Answer:

150:120

Step-by-step explanation:

U add the 2 ratios 5+4=9

270/9=30

30*5=150

30*4=120

You might be interested in

78/25 as a terminating decimal

Kazeer [188]

To write $\frac{78}{25}$ as a terminating decimal we will turn this fraction to a decimal. To do this we will divide the numerator by the deliminator and the result will be the decimal.Then we will check if it is a terminating decimal or repeating decimal. Lets do it:-

$\frac{78}{25}$
78 ÷ 25 = 3.12

3.12 is a terminating decimal.

So, $\frac{78}{25}$ as a terminating decimal is 3.12.

Hope I helped ya!!

8 0

3 years ago

Read 2 more answers

Solve this question. Check your solution. -5 + s > -1

HACTEHA [7]

−5+s>−1

Step 1: Simplify both sides of the inequality.

s−5>−1

Step 2: Add 5 to both sides.

s−5+5>−1+5
s>4

Answer:s>4

hope this helps!

3 0

3 years ago

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

Max, a carpet installer, needs to calculate the amount of carpet needed to cover a floor that’s 3 1⁄2 yards wide and 4 3⁄4 yards

AlekseyPX

Answer:

13 1/2

Step-by-step explanation:

Just multiply 3.5 and 4.75

AKA

3 1/2 and 4 3/4

Change 1/2 into 2/4 so the denominators are the same

Multiply the numerators 2 and 3, you now have 6/4

3 × 4 = 12

6/4 is improper so, you need to simplify

now you have 1 1/2

Now 12 + 1 1/2 = 13 1/12

5 0

3 years ago

Double a number and subtract nine. algebraic expression

insens350 [35]

Answer:

2y - 9

Step-by-step explanation:

number = y

2 × y - 9

2 × y can be simplified to 2y.

2y - 9

3 0

3 years ago