Ask question

Ask question

All categories

3 years ago

13

Problem

1 answer:

Nitella [24]3 years ago

8 0

Answer:maybe 29

Step-by-step explanation:

You might be interested in

Solve using the elimination method. <br> show your work <br> 2x - y = 6 <br> x + y = 6

bekas [8.4K]

Answer:

x=4, y=2

Step-by-step explanation:

2x - y = 6

x + y = 6

Add these together to eliminate y

2x - y = 6

x + y = 6

-----------------

3x = 12

Divide by 3

3x/3 = 12/3

x=4

But we still need y

x+y =6

4+y =6

Subtract 4 from each side

4+y-4 = 6-4

y=2

7 0

3 years ago

Can somebody help me with that? Thank:)

Helga [31]

The sum of the angles in a triangle is 180, so the equation will be :
2k+k+45= 180
add similar variables
3k+45=180
3k=135 (divide each side by 3)
k= 45
So 2k equals= 90
k=25
This is right angle triangle for your info!

3 0

3 years ago

Enter a recursive rule for the geometric sequence. 6, −18, 54, −162, ...

svlad2 [7]

Answer:

The required recursive formula is:

$a_n=(-3)\times a_{n-1}$

Step-by-step explanation:

We are given a geometric sequence as:

6,-18,54,-162,.....

Clearly after looking at different terms of the sequence we could observe that the sequence is an geometric progression (G.P.) with common ratio= -3 denoted by r.

Let $a_n$ represents the nth term of the sequence.

This means that:

$a_1=6, a_2=-18, a_3=54, a_4=-162,......$

As the common ratio is -3.

so,

$a_1=6\\\\a_2=-18=(-3)\times a_1\\\\a_3=54=(-3)\times a_2\\\\.\\.\\.\\.\\.\\.\\.\\.\\.a_n=(-3)\times a_{n-1}$

Hence, the required recursive formula for the geometric sequence is:

$a_n=(-3)\times a_{n-1}$

5 0

3 years ago

If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c

dolphi86 [110]

Given:

$(ax+2)(bx+7)=15x^2+cx+14$

And

$a+b=8$

Required:

To find the two possible values of c.

Explanation:

Consider

$\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}$

So

$\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}$

And also given

$a+b=8---(2)$

Now from (1) and (2), we get

$\begin{gathered} a+\frac{15}{a}=8 \\ \\ a^2+15=8a \\ \\ a^2-8a+15=0 \end{gathered}$ $a=3,5$

Now put a in (1) we get

$\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}$

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

$\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}$

When a=5 and b=3, we have

$\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}$

Final Answer:

The option D is correct.

31 and 41

8 0

1 year ago

Does anyone have answers for the Unit 6-Lesson 9: Rational Expressions and Functions Unit Test? I really need them, and will giv

GalinKa [24]

Answer:

on edge?

Step-by-step explanation:

5 0

3 years ago