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lana [24]
3 years ago
9

Need help w precal/algebra 2

Mathematics
1 answer:
Daniel [21]3 years ago
3 0

Answer:

g(-2) = -7

Step-by-step explanation:

To find g(-2), plug -2 into g(x).

g(x) = -4x² + 3x + 15

g(-2) = -4(-2)² + 3(-2) + 15

g(-2) = -4(4) + 3(-2) + 15

g(-2) = -16 + (-6) + 15

g(-2) = -1 + (-6)

g(-2) = -7

You might be interested in
QUESTION IN THE ATTACHMENT
eimsori [14]

Answer:

A. The sum of the first 10th term is 100.

B. The sum of the nth term is n²

Step-by-step explanation:

Data obtained from the question include:

Sum of 20th term (S20) = 400

Sum of 40th term (S40) = 1600

Sum of 10th term (S10) =..?

Sum of nth term (Sn) =..?

Recall:

Sn = n/2[2a + (n – 1)d]

Sn is the sum of the nth term.

n is the number of term.

a is the first term.

d is the common difference

We'll begin by calculating the first term and the common difference. This is illustrated below:

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2a + (20 – 1)d]

S20= 10 [2a + 19d]

S20 = 20a + 190d

But:

S20 = 400

400 = 20a + 190d .......(1)

S40 = 40/2 [2a + (40 – 1)d]

S40 = 20 [2a + 39d]

S40 = 40a + 780d

But

S40 = 1600

1600 = 40a + 780d....... (2)

400 = 20a + 190d .......(1)

1600 = 40a + 780d....... (2)

Solve by elimination method

Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:

40 x equation 1:

40 x (400 = 20a + 190d)

16000 = 800a + 7600. ........ (3)

20 x equation 2:

20 x (1600 = 40a + 780d)

32000 = 800a + 15600d......... (4)

Subtract equation 3 from equation 4

Equation 4 – Equation 3

32000 = 800a + 15600d

– 16000 = 800a + 7600d

16000 = 8000d

Divide both side by 8000

d = 16000/8000

d = 2

Substituting the value of d into equation 1

400 = 20a + 190d

d = 2

400 = 20a + (190 x 2)

400 = 20a + 380

Collect like terms

400 – 380 = 20a

20 = 20a

Divide both side by 20

a = 20/20

a = 1

Therefore,

First term (a) = 1.

Common difference (d) = 2.

A. Determination of the sum of the 10th term.

First term (a) = 1.

Common difference (d) = 2

Number of term (n) = 10

Sum of 10th term (S10) =..?

Sn = n/2 [2a + (n – 1)d]

S10 = 10/2 [2x1 + (10 – 1)2]

S10 = 5 [2 + 9x2]

S10 = 5 [2 + 18]

S10 = 5 x 20

S10 = 100

Therefore, the sum of the first 10th term is 100.

B. Determination of the sum of the nth term.

First term (a) = 1.

Common difference (d) = 2

Sum of nth term (Sn) =..?

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [2x1 + (n – 1)2]

Sn = n/2 [2 + 2n – 2]

Sn = n/2 [2 – 2 + 2n ]

Sn = n/2 [ 2n ]

Sn = n²

Therefore, the sum of the nth term is n²

6 0
2 years ago
What is the difference in length between the shortest and longest book from looking at picture above ?
svetlana [45]

Answer:

7 inches.

Step-by-step explanation:

Longest book length - Shortest book length

12-5

=7

7 0
3 years ago
What is the vertex form of 2x^2+10x-5
Nataly_w [17]
<span><span>Two Solutions
1. x =(10-√140)/4=(5-√<span> 35 </span>)/2= -0.458</span><span> 

2. x =(10+√140)/4=(5+√<span> 35 </span>)/2= 5.458</span></span>
7 0
3 years ago
11. What method can be used to prove these two triangles are congruent?
zubka84 [21]
This would be SAS. If that’s wrong I’m sorry but it’s the only thing that makes sense, the dashes indicate they are the same and then the circle at the point also indicates that they’re the same so you have two sides and one angle that are the same, so it should be SAS.
5 0
2 years ago
Which expression is equivalent to (x^4/3 x^2/3)^1/3?
KATRIN_1 [288]

Answer: x^{\frac{2}{3} }

Step-by-step explanation:

We have several properties of exponents in use here. The two that are used are:

(x^{a})(x^{b}) = x^{a + b} <em>(Exponents with the same base that are being multiplied together can have the exponents added)</em>

(x^{a})^{b} = x^{(a)(b)} <em>(A base raised to a power, and then raised to another power means that you can multiply the exponents to get the same result as doing inside operations and then outside operations)</em>

<em />

Let's apply it!

First, let's simplify what's inside the parenthesis.

x^{\frac{4}{3} } x^{\frac{2}{3} } <em>(Remember, they have the same base of "x", so we can add the exponents)</em>

x^{\frac{4}{3} + \frac{2}{3} } = x^{\frac{6}{3} } = x^{2}

Now we have (x^{2})^{\frac{1}{3} }. Let's use the second rule.

(x^{2})^{\frac{1}{3} } = x^{\frac{2}{3} }

Hope this helps! :^)

7 0
3 years ago
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