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3 years ago

0.608 divided by 0.395

Mathematics

1 answer:

liq [111]3 years ago

6 0

<span>0.605 divided by 0.395 = 1.53924050633</span>

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Complete the next step in solving this equation.

MatroZZZ [7]

Answer:

x =23

Step-by-step explanation:

After adding 3 to both sides of the equation, you get ; -4x + 3x + 26= 3 .Then subtract 26 on both sides of the equation, to get; -x (I.e from subtracting -4x + 3x) =3-26

-x = 3-26

-x = -23

After dividing both sides by -1, you get;

x=23

3 0

3 years ago

Element X decays radioactively with a half life of 13 minutes. If there are 530 grams of Element X, how long, to the nearest ten

Leona [35]

Answer:

$t \approx 50.4\,min$

Step-by-step explanation:

The time constant of the element X is:

$\tau = \frac{13\,min}{\ln 2}$

$\tau = 18.755\,min$

The decay function has the following form:

$\frac{m}{m_{o}} = e^{-\frac{t}{\tau} }$

$t = -\tau \cdot \ln \frac{m}{m_{o}}$

$t = -(13\,min)\cdot \ln \left(\frac{11\,g}{530\,g} \right)$

$t \approx 50.4\,min$

6 0

3 years ago

The chorus has 42 members. Which expression represents the number of ways a group of 7 members can be chosen to do a special per

Elanso [62]

42!/7!35! Is the answer you’re looking for

6 0

3 years ago

A, B, and C are points of tangency. CQ = 5, PQ = 10, and PR = 14. What is the perimeter

igomit [66]

*see attachment for diagram

Answer:

Perimeter = 38

Step-by-step explanation:

Recall: when two tangents are drawn to meet at a point outside a circle, the segments of the two tangents are congruent.

Given,

CQ = 5

PQ = 10

PR = 14

Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR

CQ = QB = 5 (tangents drawn from an external point)

BP = PQ - QB

BP = 10 - 5 = 5

BP = PA = 5 (tangents drawn from an external point)

AR = PR - PA

AR = 14 - 5 = 9

AR = RC = 9 (tangents drawn from an external point)

✔️Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR

= 9 + 5 + 5 + 5 + 5 + 9

Perimeter = 38

7 0

3 years ago

Scale drawing of the electronics board

lora16 [44]

Area of the electronic board is 9000 cm2

Step-by-step explanation:

Write ratios to represent the unknown actual length and width of the display to the scale length and width of the display.

The Electronic board always seems like a rectangular shape

Length= scale÷Actual = 5/36 = 5×36 = 180cm

$Length= scale÷Actual = 5/36 = 5×36 = 180cm$

Width=scale÷Actual = 1/50 = 1×50 = 50cm

$Width=scale÷Actual = 1/50 = 1×50 = 50cm$

Area of Rectangle = legnth × width

= 180cm × 50cm = 9000 cm2

$= 180cm × 50cm = 9000 cm2$

Area of the electronic board is 9000 cm2

6 0

3 years ago