All categories

2 years ago

Subtract 6.28-1.26 =? a)4.98 b)5.02 c)5.20 d)5.28

Mathematics

2 answers:

gladu [14]2 years ago

8 0

Answer:

b) 5.02

Step-by-step explanation:

6.28-1.26 =?

Line up the numbers,

6.28

- 1.26

Subtract left from right,

6.28

- 1.26

8 - 6

⇒ 2

6.28

- 1.26

2 - 2

⇒ 0

6.28

- 1.26

5.02

6 - 1

⇒ 5

6.28

- 1.26

5.02

Hence, the answer is 5.02, ⇒ option b

Ket [755]2 years ago

5 0

Answer:

C. 5.02

Step-by-step explanation:

You might be interested in

SOMEONE PLEASE HELP ??? Which of the following proportionality statements is correct ?

Leni [432]

Answer:

The third one is correct

Step-by-step explanation:

8 0

3 years ago

Half-life of Po-210 is 140 days. If the initial mass of the sample is 5

Anna35 [415]

Answer:

$0.625kg$

Step-by-step explanation:

we are given half-life of PO-210 and the initial mass

we want to figure out the remaining mass after 420 days

in order to solve so we can consider the half-life formula given by

$\displaystyle f(t) = a {0.5}^{t/T}$

where:

f(t) is the remaining quantity of a substance after time t has elapsed.
a is the initial quantity of this substance.
T is the half-life

since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5

thus substitute:

$\displaystyle f(420)=5\cdot{0.5}^{420/140}$

reduce fraction:

$\displaystyle f(420)=5\cdot{0.5}^{3}$

By using calculator we acquire:

$\displaystyle f(420)=0.625$

hence, the remaining sample after 420 days is 0.625 kg

4 0

3 years ago

Which of the following is 2/11 closest to? 0. 1/2. 1.

VLD [36.1K]

2/11 is 0.181818 as a decimal so closest to 0.

4 0

3 years ago

Read 2 more answers

A spinner has 10 numbered sections. The arrow on the spinner was spun 30 times, and the results are recorded in the table. Based

Ann [662]

I can’t understand the question can you write it more clearly?

7 0

3 years ago

Read 2 more answers

X2 - 6x + 12 and y = 2x - 4, algebraically are

olga_2 [115]

The solution is (4, 4)

Solution:

Given system of equations are:

$y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2$

Substitute eqn 2 in eqn 1

$x^2 - 6x + 12 = 2x - 4$

Make the right side of equation 0

$x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0$

Solve by quadratic equation

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4$

Substitute x = 4 in eqn 2

y = 2(4) - 4

y = 8 - 4

y = 4

Thus solution is (4, 4)

5 0

3 years ago