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2 years ago

Helppppp plsssssssssssssssssssssss thanks so much:D

Mathematics

2 answers:

vfiekz [6]2 years ago

8 0

Answer:

Hi, there the answer is 13pt

Hope this helps take care :)

Step-by-step explanation:

scoray [572]2 years ago

5 0

Answer:

13 pts.

Step-by-step explanation:

Hope this helps! if it does, can you mark me brianliest? I am almost to level ace!

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Factor the expression.<br> 20b + 35 =<br> (Factor completely.)

Airida [17]

Answer:

I think it would be 5(4b+7)

Step-by-step explanation:

I divided both terms by 5.

6 0

2 years ago

What is the rule for 0.7, 1.7, 2.7, 3.7

larisa86 [58]

Add 1

0.7+1=1.7

1.7+1= 2.7

etc., etc.

5 0

3 years ago

4×(2.1+3.9)-7)+6×(6.2-4.2)=

Helen [10]

I am pretty sure it is 29 because i did the math and that is what it comes out to

8 0

2 years ago

$100 is invested at 12% per year. If the amount is compounded annually, write the total amount after 2 years in exponential func

Sladkaya [172]

Answer:

A = $100(1.12)^2

Step-by-step explanation:

The standard formula for compound interest is given as;

A = P(1+r/n)^(nt) .....1

Where;

A = final amount/value

P = initial amount/value (principal)

r = rate yearly

n = number of times compounded yearly.

t = time of investment in years

For this case;

P = $100

t = 2years

n = 1

r = 12% = 0.12

Substituting the values, we have;

A = $100(1+0.12)^(2)

A = $100(1.12)^2

7 0

2 years ago

For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?

DaniilM [7]

<h2>Answer:</h2>

The values of x for which the given vectors are basis for R³ is:

$x\neq 1$

<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

$\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]$

Now, the determinant of this matrix is:

$\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1$

Hence,

$-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1$

4 0

2 years ago