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Levart [38]
3 years ago
7

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

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Compare. use <, > ,or = to complete the statement
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Pls answer 13 b ill mark u brainlist
gladu [14]

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7 students

Step-by-step explanation:

1+2+4=7students

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Solve for b. 9b = -6 + 10b
Aleksandr-060686 [28]

Answer:

b=6

Step-by-step explanation:

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

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7 0
3 years ago
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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
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