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77julia77 [94]
3 years ago
13

Factorize 2a^2+7a-15​

Mathematics
2 answers:
tresset_1 [31]3 years ago
8 0

Answer:

2a^2+10a-3a-15

2a(a+5)-3(a+5)

(2a-3) (a+5)

loris [4]3 years ago
5 0

\huge\bf  \pink {\underline{Solution :-}}

2 {a}^{2}  + 7a - 15

= 2 {a}^{2}  + 10a - 3a - 15

= 2a(a + 5) - 3(a + 5)

= (a + 5)(2a - 3)

\sf \red{Hence, Answer \:  is  \: (a + 5)(2a - 3).}

\\

\bf \purple{ \underline{ Important  \: Formulas  \: for  \: Factorization :-}}

• \:  {a}^{2}  + 2ab + {b}^{2} =  {(a + b)}^{2}

• \: (a + b)(a - b) =  {a}^{2}  -  {b}^{2}

• \:  {a}^{2} - 2 ab  +  {b}^{2}  =  {(a - b)}^{2}

• \:  {a}^{3}  + 3 {a}^{2} b + 3a {b}^{2}  +  {b}^{3} =  {(a + b)}^{3}

• \:  {a}^{3}   -  3 {a}^{2} b + 3a {b}^{2}   -  {b}^{3} =  {(a  - b)}^{3}

•  \:  {(a + b)}^{2}  +  {(a - b)}^{2}  = 2( {a}^{2}  +  {b}^{2} )

• \: {(a + b)}^{2}   -   {(a - b)}^{2}  = 4ab

• \: (a + b)( {a}^{2}   -ab   +  {b}^{2}  ) =  {a}^{3}  +  {b}^{3}

• \: (a  -  b)( {a}^{2}    + ab   +  {b}^{2}  ) =  {a}^{3}   -  {b}^{3}

• \:   {( \frac{a + b}{2} )}^{2}  - ( {\frac{a - b}{2} } )^{2}  = ab

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Factor this polynomial expression.
tankabanditka [31]

Answer:

D:  3(x + 5)(x + 5)

Step-by-step explanation:

Try multiplying the last answer choice (D); the result is 3(x^2 + 10x + 25), or

3(x + 5)(x + 5).

5 0
3 years ago
What is the value in meters per second squared of the acceleration of gravity?
Lady bird [3.3K]
In real-life physics, acceleration due to gravity will change respect to altitude. Check out the attached image (source: Wikipedia).

However, in high-school physics, acceleration due to gravity is represented by the letter g, which has a value of 9.81 meters per second squared.

This means that an object being dropped straight down will be moving 9.81 meters per second faster than it was 1 second ago. I hope this helps! :)

7 0
3 years ago
Epress 4+3i/3-4i in form of a + bi .evaluate your final answer raised to power 6​
VashaNatasha [74]

Answer:(a) Express the complex number (4 −3i)3 in the form a + bi. (b) Express the below complex number in the form a + bi. 4 + 3i i (5 − 6i) (c) Consider the following matrix. A = 2 + 3i 1 + 4i 3 − 3i 1 − 3i Let B = A−1. Find b22 (i.e., find the entry in row 2, column 2 of A−1)

Step-by-step explanation:

7 0
2 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
Point N is on line segment MO. Given NO=2x−3, MO=3x+5, and MN=2x+3, determine the numerical length of MO.
ELEN [110]

Answer:-19

Step-by-step explanation:

3x+5-(2x-3)=0

3x+5-2x+3=0

x+8=0

x=-8

substitute x into 3x+5 to find the length of MO

3x+5

3(-8)+5

-24+5

-19

6 0
3 years ago
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