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MAXImum [283]
2 years ago
10

Factor (5a–3b)^2–25a^2 PLS HELP 50 POINTS

Mathematics
2 answers:
Nonamiya [84]2 years ago
3 0

Answer:

(-3b) (10a -3b)

Step-by-step explanation:

(5a–3b)^2–25a^2

Replace (5a–3b) with x

x^2 - (5a) ^2

We have the difference of squares

( x-5a) (x+5a)

Now replace x with (5a -3b)

(5a -3b -5a) ( 5a -3b +5a)

Combine like terms

(-3b) (10a -3b)

KatRina [158]2 years ago
3 0

Answer:

(5a-3b)²-25a²

= 25a²+9b²-2×5a×3b-25a { (a-b)²= a²+b²-2ab}

=9b²-30ab

=3b(3b-10a)

= -3b(10a-3b)

Ans

Hope, it helped!

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Tim and Jim are working on a school project together. If he had to work by himself, it would take
Vladimir [108]

Answer: 6 hours

Step-by-step explanation:

Given, Tim and Jim are working on a school project together.

Tim takes 10 hours to complete the school project.

If Jim worked alone, it would take him 15 hours

Let t be the time they both take if they work together.

Then, \dfrac{1}{t}=\dfrac{1}{\text{Time taken by Tim}}+\dfrac{1}{\text{Time taken by jim}}

\Rightarrow\ \dfrac{1}{t}=\dfrac{1}{10}+\dfrac{1}{15}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{3+2}{30}=\dfrac{5}{30}\\\\\Rightarrow\dfrac{1}{t}=\dfrac{1}{6}\\\\\Rightarrow t=6

So, it will take<u> 6 hours</u> to complete the school project together .

4 0
2 years ago
As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur
Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math.  It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

y=Ce^{kt}

Filling in our formula with the 2 conditions we are given:

65=Ce^{10k}   and   85=Ce^{15k}

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k.  If

65=Ce^{10k} then

\frac{65}{e^{10k}}=C which, by exponential rules is the same as

C=65e^{-10k}

Since that value of C is the same as the value of C in the other equation, we sub it in:

85=65e^{-10k}(e^{15k})

Divide both sides by 65 and use the rules of exponents again to get

\frac{85}{65}=e^{-10k+15k} which simplifies down to

\frac{85}{65}=e^{5k}

Take the natural log of both sides to get

ln(\frac{85}{65})=5k

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

65=Ce^{10(.0536527973)}

which simplifies down to

65=Ce^{.536527973}

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

y=38.01038064e^{(.0536527973)(23)} which simplifies a bit to

y=38.01038064e^{1.234014338}

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565

6 0
3 years ago
Solve for x please help ! (show work)
Alina [70]

Answer:

x = -5

Step-by-step explanation:

-(5x-2) = 27

Distribute the minus sign

-5x +2 = 27

Subtract 2 from each side

-5x +2-2 = 27-2

-5x = 25

Divide by -5

-5x/-5 = 25/-5

x = -5

5 0
3 years ago
Read 2 more answers
Raise <br> n<br> to the 6th <br> power,<br> then add 5 to the result
aleksley [76]
Except there are values you want to substitute into the expression, this is what your answer looks like:

n^{6}+5
7 0
3 years ago
A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean
elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0
3 years ago
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