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3 years ago

Factor the Trinomial:2a^2 + 3a + 1

Mathematics

2 answers:

ohaa [14]3 years ago

8 0

2a^2 + 3a + 1

= 2a^2 + 2a + a + 1

= 2a(a + 1) + 1 (a + 1) ( a + 1) is common)) so it is

= (2a + 1)(a + 1) (answer)

yKpoI14uk [10]3 years ago

5 0

Try "factoring by grouping."

The binomial factors of 2a^2 + 3a + 1 must start with a and 2a respectively, so that the product of these first terms is 2a^2. Also note that a and 2a must add up to 3a. They do.

Then 2a^2 + 3a + 1 = (2a+1)(a+1).

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A plate of glass with parallel faces having a refractive index of 1.57 is resting on the surface of water in a tank. A ray of li

frutty [35]

Answer:

$27.26^{\circ}$

Step-by-step explanation:

We are given that

$n_g=1.57$

$n_w=1.33$

Angle of incidence,i=37.5 degree

We have to find the angle of refraction into the water made by ray wit the normal to the surface.

Refractive index of water, $n_a=1$

By Snell's law

$\frac{n_g}{n_a}=\frac{sin i}{sin r}$

$sin r=\frac{n_a sini}{n_g}=\frac{1\times sin37.5}{1.57}$

$sin r=0.388$

$r=sin^{-1}(0.388)=22.83^{\circ}$

Again for glass water surface

Using Snell's law

Angle of incidence,i'=22.83 degree

$\frac{n_w}{n_g}=\frac{sin i}{sin r}$

$\frac{1.33}{1.57}=\frac{sin22.83}{sin r'}$

$sin r'=\frac{1.57sin(22.83)}{1.33}=0.458$

$r'=sin^{-1}(0.458)=27.26^{\circ}$

8 0

3 years ago

Are these ratios equivalent?<br> 12 cards to 72 animals, 11 marbles to 66 marbles

Dmitrij [34]

yes

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66/11 = 6

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3 years ago

Solve for u. -14 = u ÷ 8

Elodia [21]

The correct answer is

u= -112

6 0

3 years ago

Last one - Thank you an advance :)

Ivan

Answers:
a. Given
b. Transitive property of congruence
c. Vertical angles are congruent
d. Transitive property of congruence

Let me know if you need any clarification as to how I got those answers. They should be self-explanatory but I'm happy to clarify further if needed.

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3 years ago

Read 2 more answers

Plz solve this problem of trigonometry<br>i am an aakashian

makkiz [27]

Step-by-step explanation:

$\bf L.H.S = \tt \dfrac{sec\: \theta + tan \: \theta - 1}{tan \: \theta - sec \: \theta + 1} \\ \\$

$: \implies \tt \dfrac{\frac{1}{cos \: \theta} + \frac{sin \: \theta}{cos \: \theta} - 1}{ \frac{sin \: \theta}{cos \: \theta} - \frac{1}{cos \: \theta} + 1 } \: = \dfrac{1 + sin \: \theta - cos \: \theta}{sin \: \theta + cos \: \theta} \\ \\$

$: \implies \tt\dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta + (cos \: \theta - 1)} \: \times \: \dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta - (cos \: \theta - 1)} \\ \\$

$: \implies \tt\dfrac{ sin^{2} \: \theta + cos^{2} \: \theta + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: (cos \: \theta - 1)}{sin^{2} \: \theta - (cos \: \theta - 1)^{2} } \\ \\$

$: \implies \tt\dfrac{1 + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - 1 + 2 \: cos \: \theta } \\ \\$

$: \implies \tt\dfrac{2 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - sin^{2} \: \theta - cos^{2} \: \theta + 2 \: cos \: \theta } \\ \\$

$: \implies \tt\dfrac{2 (1 - \: cos \: \theta )- 2 \: sin \: \theta (1 - \: cos \: \theta)}{ 2 \: cos \: \theta - 2 \: cos^{2} \: \theta} \\ \\$

$: \implies \tt\dfrac{(2 + 2 \: sin \: \theta) \: \cancel{(1 - cos\: \theta)}}{2 \: cos \: \theta \: \cancel{(1 - cos \: \theta)}} \: = \: \dfrac{1 + sin \: \theta}{cos \: \theta} \\ \\$

$: \implies\tt\dfrac{1 + sin \: \theta}{cos \: \theta} \: \times \: \dfrac{1 - sin \: \theta}{1 - sin \: \theta} \\ \\$

$: \implies\tt\dfrac{1 + sin^{2} \: \theta}{cos \: (1 - sin \: \theta)} \\ \\$

$: \implies\tt\dfrac{cos^{2} \: \theta}{cos \: \theta (1 - sin \: \theta)} \\ \\$

$: \implies\tt\dfrac{cos \: \theta}{1 - sin \: \theta} \: = \: \bf{ R.H.S}\\ \\$

$\huge\bigstar \:\underline{\red{\sf Hence, Proved}} \: \bigstar \\$

6 0

4 years ago