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umka2103 [35]
3 years ago
10

Susan incorrectly factored the expression below. 12a − 15b + 6 3 (4a + 5b + 3)

Mathematics
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

3(4a−5b+2)

Step-by-step explanation:

The answer should have originally been 3(4a-5b+2)

3(4a+5b+3)= 12a+15b+9 which is incorrect.

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Evaluate -6(x-2y) when x= -3 and y= -5<br> a. -42<br> b. 80<br> c. -80<br> d. -16
Vinil7 [7]
Simply substitute x for -3 and y for -5, creating the equation shown below:
     
     -6 (-3 - 2(-5) )
Now that you have your values set up, now you can do the math by starting off with Multiplying -2 by -5, which will get you 10. 

Now that you have the parenthesis simplified in the parenthesis, you can use the distributive property by multiplying all the numbers with -6, like so:
-6(-3+10)= 18 - 60, which simplifies to -42, in which (A) is your answer.
 
 
5 0
4 years ago
Read 2 more answers
Please answer CORRECTLY! Will give brainliest as well as 55 points. (XD YES, I am playing Prodigy even though I am an eight grad
chubhunter [2.5K]

Answer:

60

Step-by-step explanation:

4 0
3 years ago
Find a1 if Sn = 89,800, r = 3.4, and n = 10. Round to the nearest hundredth if necessary.
xxMikexx [17]

Answer:

a_1=1.04

Step-by-step explanation:

We have a geometric  sequence with:

Sn = 89,800, r = 3.4, and n = 10

Where

Sn is the sum of the sequence

r is the common ratio

a_1 is the first term in the sequence

n is the number of terms in the sequence

The formula to calculate the sum of a finite geometric sequence is:

S_n=\frac{a_1(1-r^n)}{1-r}

Then:

89,800=\frac{a_1(1-(3.4)^{10})}{1-3.4}

Now we solve for a_1

89,800(1-3.4)=a_1(1-(3.4)^{10})

a_1=\frac{89,800(1-3.4)}{1-(3.4)^{10}}\\\\a_1=1.04

4 0
4 years ago
Sinx =căn 2/3<br> sinx=5/4<br> sinx =1<br> sin3x=can3/2<br> sin(x-60)=-1/2<br> sin3x=1/2
mel-nik [20]

Answer:

Correct option is

B

0

D

−1

sinx+sin2x+sin3x

=sin(2x−x)+sin2x+sin(2x+x)

=2sin2xcosx+sin2x [ by using sin(A+B)=sinAcosB+sinBcosA and sin(A−B)=sinAcosB−sinBcosA ]

=sin2x(2cosx+1)........(i)

cosx+cos2x+cos3x

=cos(2x−x)+cos2x+cos(2x+x)

=2cos2xcosx+cos2x [By using cos(a−b)=cosa⋅cosb+sina⋅sinb and cos(a+b)=cosa⋅cosb−sina⋅sinb]

=cos2x(2cosx+1).....(ii)

∴(sinx+sin2x+sin3x)

2

+(cosx+cos2x+cos3x)

2

=1

sin

2

2x(2cosx+1)

2

+cos

2

2x(2cosx+1)

2

=1.......[From(i)(ii)]

⇒(2cosx+1)

2

=1

⇒2cosx+1=±1

∴cosx=0or−1

6 0
3 years ago
What's the answer to this question?
Usimov [2.4K]
What book is that from
6 0
4 years ago
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