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3 years ago

Can you answer this please

Mathematics

1 answer:

Flauer [41]3 years ago

5 0

Look at the attachment!

Thanks. (I can’t see the Tuesday 24 Question 5, sorry.)

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Sketch the graph for each function. Choose either A, B, C, or D.

Serjik [45]

Hi. Here it is!
I hope this help...

—> And the answer is letter A

6 0

2 years ago

You have $10.50 to spend at the bowling alley. It costs $3 to rent shoes

Oksana_A [137]

Answer: 3 games

Step-by-step explanation:

10.5

First subtract 3 because you will need shoes.

7.50

Now divide by 2.50

7.5/2.5

3 GAMES

3 0

3 years ago

What is the logarithmic form of the solution to 102t = 9?

Harman [31]

Equivalent equations are equations that have the same value

The equation in logarithmic form is $t = \frac{\log(9)}{2}$

<h3>How to rewrite the equation</h3>

The expression is given as:

$10^{2t} = 9$

Take the logarithm of both sides

$\log(10^{2t}) = \log(9)$

Apply the power rule of logarithm

$2t\log(10) = \log(9)$

Divide both sides by log(10)

$2t = \frac{\log(9)}{\log(10)}$

Apply change of base rule

$2t = \log_{10}(9)$

Divide both sides by 2

$t = \frac{\log_{10}(9)}{2}$

Rewrite as:

$t = \frac{\log(9)}{2}$

Hence, the equation in logarithmic form is $t = \frac{\log(9)}{2}$

Read more about logarithms at:

brainly.com/question/25710806

8 0

2 years ago

How much carbon dioxide will you prevent from entering the atmosphere in one year, if you replace a 100-watt light bulb with a C

Afina-wow [57]

Answer:

280000lb

Step-by-step explanation:

In this question, we shall be calculating the amount of carbon iv oxide prevented from entering the atmosphere.

Generally we know that 1kwh produces 1.37 lbs of Carbon iv oxide in the atmosphere

Now, using a 100 watt light bulb daily for 8 hours, for a year, the amount of Carbon iv oxide produced will be;

100 * 1.37 * 8 * 365 = 40040 lb

Using a 30 watt CFL for 8hrs/ day for a year, we have;

30 * 1.37 * 8 * 365 = 120012 lb

Now calculating the amount of CO2 prevented, we have 40040 - 120012 = 280028 lb = 280000 after rounding up

3 0

3 years ago

Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac

melomori [17]

Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ $b+c=20$
Equation [2] ⇒ $1.50b+5c=54.50$

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; $b$ and $c$ , that we need to solve, then this becomes a simultaneous equation problem.

First, rearrange Equation [1] to make either $b$ or $c$ the subject:
$b+c=20$
$b=20-c$

Then we substitute $b=20-c$ into Equation [2]
$1.50b+5c=54.50$
$1.50(20-c)+5c=54.50$
$30-1.50c+5c=54.50$
$5c-1.5c=54.50-30$
$3.5c=24.50$
$c=7$

Now we know the value of $c$ which is $c=7$ , substitute this value into $b=20-c$ we have $b=20-7=13$

Answer:
Numbers of books = 13
Numbers of CDs = 7

5 0

3 years ago