All categories

3 years ago

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

You might be interested in

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

the 17th term of an arithmetic sequence is 51.

i.e. $a_{17}=51$

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

$a_{17}=a_1+\left(17-1\right)d$

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

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

$a_1=-61$

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