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4 years ago

I WILL AWARD BRAINLIEST!

Mathematics

2 answers:

sammy [17]4 years ago

8 0

A)70 this is the answer

gavmur [86]4 years ago

7 0

Answer:

A. 70

Step-by-step explanation:

You can use the transversals and use vertical angles are congruent.

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Bridget reads 5 pages of her book per day 4 days per week for several weeks.

Semenov [28]

It would be w(5•4) because she reads 20 pages in one week. then you can multiply 20 by any number of weeks to get that number of pages in total.

8 0

3 years ago

find the common difference and the three terms in the sequence after the last one given -39 -33 -27 -21

Serggg [28]

Answer:

Common difference:6 3 terms: -15, -9. -3

Step-by-step explanation:

-39, -33, -27, and -21 all have a difference of 6.

6 0

3 years ago

You cut grass with a friend for 5 hours and each got paid 8 dollars an hour. On your way home you stopped by a convenient store

jeka94

Answer:

31.50

Step-by-step explanation:

I added 8 +8+8+8+8 got 40 then subtracted 8.50

4 0

3 years ago

2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA

professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

$A=45e^{-0.0045(2500)}$ $A=45e^{-11.25}$ $A=45\times0.000013=0.000585$ $A=0.000585$

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0

1 year ago

URGENT! WILL GIVE BRAINLIEST!!! IF CORRECT!

In-s [12.5K]

Answer: $6.75

Step-by-step process:
100%-25%=75%
therefore, they will cost 75% of the original price
75% of $9.00 = .75*9
= $6.75

5 0

3 years ago

Read 2 more answers