All categories

3 years ago

Please help i have a c

Mathematics

2 answers:

tensa zangetsu [6.8K]3 years ago

8 0

Answer:

D. R is in the 4 quadrant

Sergeu [11.5K]3 years ago

6 0

Answer: the last option

Step-by-step explanation:

the top right is I, the top left is II, the bottom left is III, and the bottom right is IV

You might be interested in

Compare. use <, > ,or = to complete the statement

8_murik_8 [283]

> hello there stranger

8 0

3 years ago

Pls answer 13 b ill mark u brainlist

gladu [14]

Answer:

7 students

Step-by-step explanation:

1+2+4=7students

5 0

2 years ago

Solve for b. 9b = -6 + 10b

Aleksandr-060686 [28]

Answer:

b=6

Step-by-step explanation:

9b-10b=-6+10b-10b

-b=-6 so b=6

7 0

3 years ago

Read 2 more answers

In what time will Rs. 5600 amount to Rs. 6720 at 8% per annum?

Leona [35]

Answer: 2 years

Step-by-step explanation:

If Interest= Principal x rate x time/100; then time is (100 × Interest)/(Principal × Rate)

From the question, old Principal is 5600, new principal =6720, the difference is the interest accrued.

Interest = 6720 - 5600 = Rs 1120

Rate = 8% = 8 ÷ 100 = 0.08

Time = x

Then, slot the values into the formula

Time= 100 x 1120 / 6720 x 8

= 112000/53760

=2.08

Time= 2years

I hope this helps

8 0

3 years ago

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0

2 years ago