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3 years ago

5 = 1 and 5 in a fraction and w

Mathematics

2 answers:

alekssr [168]3 years ago

8 0

I think that the answer is 25

vovangra [49]3 years ago

7 0

What we want to do first is isolate the variable. To do this we need to multiply each side by 5 to remove the fraction

5 * 5 = 5* $\frac{1}{5}$ w
Simplify the right side
5 * $\frac{1}{5}$ w = 1w
Now simplify the left side
5 * 5 = 25
Then we get...
25 = w

There's your answer!

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Figures ABCD and PQRS are parallelogram find x and y (lengths are in metres) Part i and ii.

STatiana [176]

Step-by-step explanation:

For 1st question

In a parm ABCD

5x - 4 = 16 ( being opposite sides of parallelogram)

5x = 16 + 4

5x = 20

x = 20 / 5

x = 4

3y + 1 = 22 (being opposite sides of parallelogram)

3y = 22 -1

3y = 21

y = 21 / 3

y = 7

For second question

In parm PQRS

Diagonals bisect each other so

x + 4 = 12

x = 12 - 4

x = 8

x + y = 32

8 + y = 32

y = 32 - 8

y = 24

Hope it will help :)❤

8 0

2 years ago

(2.75 x 107) + (1.23 x 108) simplify

Thepotemich [5.8K]

Answer:

427.09

Step-by-step explanation:

(2.75 x 107) + (1.23 x 108)

294.25 + 132.84 = 427.09

6 0

2 years ago

I need help with number 2 please.

zalisa [80]

The answer is 73.3333333 ounces<span>
</span>

5 0

2 years ago

Read 2 more answers

Complete the two-column proof by providing a reason for each of the five statements.

mrs_skeptik [129]

Answer:

Reason:

1) assumptions given

2) RUTS is a rectangle

3) RUTS is a rectangle

4) because:

+) ∠STU and ∠SRU are the right angles

+) RU = ST; UT = RS

=> ΔURS ≅ ΔSTU (SAS)

5) because ΔURS ≅ ΔSTU (SAS) => ∠USR = ∠SUT

Step-by-step explanation:

5 0

2 years ago

A special liquid is held in a tank described as (x 2 + y 2 ) ≤ z ≤ 1 in a Cartesian coordinate system. Assume that the density o

vivado [14]

Since $\rho=\dfrac mV$ (density = mass/volume), we can get the mass/weight of the liquid by integrating the density $\rho(x,y,z)$ over the interior of the tank. This is done with the integral

$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$

which is more readily computed in cylindrical coordinates as

$\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}$

7 0

2 years ago