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3 years ago

I really need help on this problem can i get help on how to solve the problem please

Mathematics

1 answer:

Jobisdone [24]3 years ago

7 0

You cannot assume that the angles are equal. there is no angle measure and nothing to say the lines are parellel.

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The weight of eggs from a local farm are Normally

SVEN [57.7K]

Answer:

65.5

Step-by-step explanation:

I just took the assignment on edge. Good luck

5 0

3 years ago

Joyce Mesnic bought an HP laptop computer at Staples for $699. Joyce made a $100 down payment and financed the

s2008m [1.1K]

699-100=$599
599/12= $49.92

3 0

3 years ago

The 7th grade class experienced a 15% decrease in enrollment. Last year, the 7th grade class had 280 students. How many students

Monica [59]

238 students are enrolled in the 7th grade class this year

<u>Solution: </u>

Given that

Percentage decrease in number of students enrolled in 7th grade = 15%

Number of students enrolled last year = 280

Need to determine number of students enrolled in 7th grade this year.

Lets assume number of students enrolled this year = x

<em>Percentage decrease in enrollment </em>= (Number of student enrolled last year – Number of students enrolled this year) whole divided by number of students enrolled last year multiplied by 100

$\begin{array}{l}{\Rightarrow 15=\frac{280-x}{280} \times 100} \\\\ {=>15 \times 280=28000-100 x} \\\\ {=>100 \mathrm{x}=28000-4200=23800} \\\\ {\Rightarrow x=\frac{23800}{100}} \\\\ {\Rightarrow x=238}\end{array}$

Hence number of students enrolled this year = 238.

7 0

3 years ago

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours

ella [17]

Answer:

137

Step-by-step explanation:

Use the model A = Pe^(kt), where P is the initial number and k is the growth constant. Then, in case 1,

300 = Pe^(k*3)

and in case 2,

4000 = Pe^(k*10). We need to find P and e^k.

300 = Pe^(k*3) can be rewritten as (300/P) = [e^k]^3, which in turn may be solved for e^k:

∛(300/P) = e^k

Now we go back to 4000 = Pe^(k*10) and rewrite it as (4000/P) = [e^k]^10. Substituting ∛(300/P) for e^k, we obtain:

(4000/P) = (300/P )^(10/3)

which must now be solved for P. Raising both sides by the power of 3, we get:

(4000/P)^3 = (300/P)^10. Therefore,

4000^3 300^10

------------- = -------------

P^3 P^10

which reduces to:

4000^3 300^10

------------- = -------------

1 P^7

------------ = -----------

4000^3 300^10

or:

300^10

P^7 = ---------------- = 9.226*10^13 = 92.26*10^12

4000^3

Taking the 7th root of both sides results in P = (92.26*10^12)^(1/7), or

P = 137.26

The initial number of bacteria was 137.

7 0

3 years ago

Lily deposited $500 in a savings account. At the end of 1 year, she had earned $20 in simple interest. What interest rate did sh

rjkz [21]

X/100 *500 = 20

500/100x = 20

5x = 20

x = 4 %

7 0

3 years ago

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