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steposvetlana [31]
3 years ago
8

I give brainliest!! no link or i’ll report you.

Mathematics
2 answers:
Yuki888 [10]3 years ago
5 0
The answer might be 13 or 25
umka2103 [35]3 years ago
3 0
The answer is 13 units
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A woman has a total of $ 12 , 000 to invest. She invests part of the money in an account that pays 7 % per year and the rest in
arlik [135]
That would be 34 your welcome:)
3 0
3 years ago
The 17th term of an Arithmetic sequence is 51. If the commons difference is 7, what is the first term ?
vovikov84 [41]

Answer:

The first term is:

  • a_1=-61

Step-by-step explanation:

The arithmetic sequence is defined by

a_n=a_1+\left(n-1\right)d

where

a_1 is the first term

d is the common difference

as

the 17th term of an arithmetic sequence is 51.

i.e. a_{17}=51

so

a_n=a_1+\left(n-1\right)d

a_{17}=a_1+\left(17-1\right)d        

51=a_1+\left(17-1\right)7         ∵ n = 17, a_{17}=51 , d=7

a_1+112=51                   ∵ \left(17-1\right)\cdot \:\:7=112

a_1=-61

Therefore, the first term is:

  • a_1=-61
8 0
3 years ago
In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.
DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2

                           + 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5

5C_0(\frac{1}{ax})^5(2ax^2)^0  =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1  = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\

5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

6 0
3 years ago
What is the volume of a right rectangular prism with a length of 19 centimeters, a width of 3 6 centimeters, and a height of 2.7
solong [7]

Answer:

184.68cm

Step-by-step explanation:

19x3.6=68.4

68.4x2.7=184.68

I think the answer is C. and you accidentally put the dot in the wrong place.

7 0
3 years ago
I WILL GIVE BRAINIEST!
Novay_Z [31]
Range is the value of y
f(x)=-x²+5
when x=-1, f(x)=-(-1)²+5=4
when x=2, f(x)=-(2)²+5=1
so 1<y≤4
the second piece: f(x)=2x-3
when x=2, f(x)=2(2)-3=1
when x=5, f(x)=2(5)-3=7
so 1≤y<7

union of the two parts of y: 1≤y<7
[1,7) is the answer. 
3 0
3 years ago
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