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quester [9]
3 years ago
7

What is the nth term for 3,4.5,6,7.5

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
8 0
If nth means ninth then the answer is 15.  If nth is not ninth then tell me I am glad to help!
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Please help me please please help please
Andre45 [30]

Answer:

Step-by-step explanation:

6 0
3 years ago
Find the area of the trapezoid.<br> bı = 5, b2 = 7, h = 4<br> The area is<br> square units
zmey [24]

Answer:

24 square units

Step-by-step explanation:

8 0
3 years ago
What are the first 6 multiples of 3?
zvonat [6]

Answer:

1 and 3 so 3

Step-by-step explanation:

6:1,2,3,6

3:1,3

7 0
3 years ago
Oscar wants to rent a boat. Boats 4 Less charges a fee of $100 and $35 per hour after that. Boats-R-Us doesn't charge a fee but
Reika [66]

The rental company to hat Oscar should use is the second company because it's cheaper.

<h3>How to illustrate the information?</h3>

The total amount charged for the first boat will be:

= 100 + 35h

where h = number of hours

= 100 + 35(8)

= 100 + 260

= $360

The amount charged for the second company will be:

= 50 × 8

= 400

Therefore, the rental company to hat Oscar should use is the second company because it's cheaper.

Learn more about equations on:

brainly.com/question/2972832

#SPJ1

7 0
1 year ago
please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it
katovenus [111]

Since \alpha and \beta are roots of ax^2+bx+c, we can factorize the quadratic in terms of the roots as

ax^2+bx+c = a(x-\alpha)(x-\beta)

Expanding the right side gives

ax^2+bx+c = ax^2-a(\alpha+\beta)x+a\alpha\beta

so that

\alpha + \beta = -\dfrac ba \\\\ \alpha\beta = \dfrac ca

A polynomial with roots \alpha^2 and \beta^2 would be

(x-\alpha^2)(x-\beta^2)

and expanding this gives

x^2 - (\alpha^2+\beta^2)x+\alpha^2\beta^2

Now,

\alpha\beta = \dfrac ca \implies \alpha^2\beta^2 = (\alpha\beta)^2 = \dfrac{c^2}{a^2}

and

(\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \\\\ \implies \alpha^2+\beta^2 = \left(-\dfrac ba\right)^2 -2\left(\dfrac ca\right) = \dfrac{b^2-2ac}{a^2}

So we can write the second quadratic in terms of a,b,c as

(x-\alpha^2)(x-\beta^2) = x^2 - \dfrac{b^2-2ac}{a^2}x + \dfrac{c^2}{a^2}

and to make things look cleaner, scale the whole expression by a^2 to get

\boxed{a^2x^2 + (2ac-b^2)x + c^2}

4 0
3 years ago
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