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2 years ago

I need some help...

1 answer:

7 0

-------------------------------------
Question
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17x³ + (3x + 8x³)

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Remove the bracket
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17x³ + 3x + 8x³

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Combine like terms
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25x³ + 3x

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Take out common factor
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x(25x² + 3)

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Answer: x(25x² + 3)
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You might be interested in

Select all that apply

katovenus [111]

The range is Y>2. :)

5 0

2 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

2 years ago

A=1/2bh. A=200ft. H=10. What is b

Svetllana [295]

Answer:

B= $\frac{2A}{H}$

Step-by-step explanation:

If you noticed, A= $\frac{1}{2}$ BH is the formula for finding the area for a triangle. Your goal is to get B by it's self. Your first step will be to clear of the fraction first, so you will multiply both sides by 2. 2(A)=2( $\frac{1}{2}$ BH). On the left, you have 2×A= On the right side, you have 2( $\frac{1}{2}$ BH), but since you have a number in the equation, you will only use 2× $\frac{1}{2}$ . To solve 2× $\frac{1}{2}$ , you will cnacel out both 2's and you have 1. 1×BH will still equal BH, so you are now left will B×H.

(Your new equation looks like this by the way). 2A=BH

Since you need to get B by its self, the way to clear the H away from the B is by dividing. You will now divide the B and H aswell as 2A and H. (It will look like this) $\frac{2A}{H} = \frac{BxH}{H}$ . (Again when you have the same number or letter, you cross it out. When you divide, you won't change anything on the left side, and all you have to do on the right id to cross out the H next to the B and cross out the H on the bottom of the equation). You should be left with $\frac{2A}{H}$ = B. Now you can turn it around for your final answer. B= $\frac{2A}{H}$ .

Please let me know if i helped, how I did, and if you have any questions.

3 0

2 years ago

Brooke's bedroom has an area of 36 square feet. The bedroom is 6 times as many square feet as Brooke's closet. If the closet is

denis23 [38]

Answer:

195

Step-by-step explanation:

Because 36 x 6 = 192 + 3 = 195

6 0

2 years ago

Explain why the graph of quadratic function could not contain both a minimum vertex and a maximum vertex at the same time.

KonstantinChe [14]

A quadratic function is a function of the form $f(x)=ax^2+bx+c$ . The vertex, $(h,k)$ of a quadratic function is determined by the formula: $h= \frac{-b}{2a}$ and $k=f(h)$ ; where $h$ is the x-coordinate of the vertex and $k$ is the y-coordinate of the vertex. The value of $a$ determines if the parabola opens upward or downward; if $a$ is positive, the parabola opens upward and the vertex is the minimum value, but if $a$ is negative the graph opens downward and the vertex is the maximum value. Since the quadratic function only has one vertex, it could not contain both a minimum vertex and a maximum vertex at the same time.

3 0

2 years ago