Factor the following.

Mathematics

1 answer:

a_sh-v [17]4 years ago

3 0

The answer would be (y + 14) (y - 8) because when multiplied it gives you
y^2 + 6y - 112

Hope this helps :)
Please give brainliest

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Answer?! Thanks I appreciate it

LenKa [72]

6.02 - 3.9 = 2.12 km
Convert 2.12 km to meters.
2.12 km = 2,120 meters.

4 0

3 years ago

30 points.. Please Help! I don't understand what to do with this question...a bit confused.

Romashka-Z-Leto [24]

Eh, here's my best try for a table:

Like tv don't like tv total
Like reading: 70 10 80
Don't like read. 20 30 50 
total 90 40 130

The percentage of students that like tv as well as reading is- 70 divided by 130 which gives you .538 which in percentage is 53.8%.

The probability of a student that does not prefer to watch tv as also does not like to spend their time reading is 30 over 40 which can be simplified or reduced to 3 over 4. [30/40 reduced to 3/4]

Hope this somehow helped you :)

5 0

3 years ago

-9 + 1/4x - 6 - 1/8x

Deffense [45]

Answer:

Assuming you want me to simplify your answer, the answer to your question is x/8-15.

Step-by-step explanation:

Simplify each term by combining 1/4 and x, and then combine x and 1/8 to get -9+x/4-6-x/8. Subtract 6 from -9 for x/4-15-x/8. Write x/4 as a fraction with a CD (Common Denominator) by multiplying by 2/2. Write each expression with a CD of 8. Combine the numerators, subract x from x*2 by reordering x and 2 to get 2*x-x/8-15. Then subract x from x to get x/8-15.

5 0

3 years ago

Read 2 more answers

Find p(0), p(1) and p(2) for the polynomial P(t) = 2 + t + 2t²-t³

FinnZ [79.3K]

p(0)=2

p(1)=2+1+2-1=4

p(2)=2+2+8-8=4

5 0

3 years ago

Find a power series representation of e^x sin(x)

liberstina [14]

The Taylor series is defined by:
$f(x) = \sum \frac{f^n (a)}{n!} (x-a)^n$

Let a = 0.
Then its just a matter of finding derivatives and determining how many terms is needed for the series.

Derivatives can be found using product rule:
$f^n (x) = e^x g^{n-1} (x) + e^x g^n (x) \\ g^0 (x) = sin x$

Do this successively to n = 6.
$f^1 (x) = e^x (sinx +cos x) \\ f^2 (x) = e^x (2cos x) \\ f^3 (x) = e^x(2cos x - 2sin x) \\ f^4 (x) = e^x (-4 sin x) \\ f^5 (x) = e^x(-4sinx -4cos x) \\ f^6(x) = e^x(-8cos x)$

Plug in x=0 and sub into taylor series:
$e^x sin x = x+x^2 +\frac{x^3}{3} -\frac{x^5}{30}-\frac{x^6}{90}...$

If more terms are needed simply continue the recursive derivative formula and add to taylor series.

6 0

4 years ago