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3 years ago

9

Please help me with this question:)!

1 answer:

Georgia [21]3 years ago

6 0

Answer:

i think so option b is the correct option.

Step-by-step explanation:

because 3+7=10 not 11.

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Is college Algebra hard ?

Degger [83]

It depends on what level of classes you take, but algebra classes in college do tend to be harder than algebra classes in high school or middle school.

7 0

3 years ago

What is the degree of the polynomial?

neonofarm [45]

Answer:

5

Step-by-step explanation:

6x^2 + 7x^3 + 8x^5 +9

Write in order from highest power to lowest power

8x^5 + 7x^3 + 6x^2 +9

The highest power ( largest exponent) is 5

The degree of the polynomial is 5

6 0

2 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago

Put the equation 3x - y > 9 into slope-intercept form (y = mx + b) but keep the inequality. (Don't change the sign to equals

Anni [7]

Answer:

y greater . ...3x-9

Step-by-step explanation:

ddsthykfdsfudfd

7 0

3 years ago

Read 2 more answers

Simplify -(x-2y)-y I need help

tatiyna

The simplified version of this equation would be

-x + y

3 0

3 years ago