All categories

3 years ago

What 2 decimals make 5.036

Mathematics

2 answers:

salantis [7]3 years ago

8 0

2 is the tens position meaning 2 tens

I am Lyosha [343]3 years ago

5 0

2.518+2.518 is equal to 5.036

You might be interested in

I dont know how to do this one :(((

MaRussiya [10]

Answer:

$y=1.8$

Step-by-step explanation:

You can compare the equation to slope intercept form where b = mid-line

$f(x)=5.6 sin(1.2x-2.3)+1.8$

$y=mx+b$

5.6 ⇒ m

$sin(1.2x-2.3)$ ⇒ x

1.8 ⇒ b

~Hope this helps!~

3 0

4 years ago

Help me!!!!!!!!!!!!!!

MA_775_DIABLO [31]

11. sin A = 7/25

cos A = 24/25

tan A = 7/24

3 0

4 years ago

What is the axis of symmetry for f(x)=2x2+8x+8

spin [16.1K]

2x^2 + 8x + 8
= 2(x^2 + 4x) + 8
= 2 [ (x + 2)^2 - 4) + 8

axis of symmetry is x = -2 ( because of the (x + 2)^2 )

8 0

3 years ago

Read 2 more answers

Rewrite this expression using a single exponent 10^2/10^5

andre [41]

Answer:

10^-3

Step-by-step explanation:

so if you have 10^2/10^5, you would subtract the exponents since it is division. 2-5=-3

therefore, it would be 10^-2

Hope this helps!

6 0

3 years ago

Use Cramer’s rule to solve for x: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17

V125BC [204]

Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

$\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}$

Cramer's rule says that

$x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}$

where $x_i$ is the solution for i-th variable, and $\mathbf A_i$ is a modified version of $\mathbf A$ with its i-th column replaced by $\mathbf b$ .

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

$\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}$

$\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}$

$\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))$

$\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120$

The modified matrices and their determinants are

$\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240$

$\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360$

$\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480$

Then by Cramer's rule, the solution to the system is

$x = \dfrac{-240}{-120} \implies \boxed{x = 2}$

$y = \dfrac{360}{-120} \implies \boxed{y = -3}$

$z = \dfrac{-480}{-120} \implies \boxed{z = 4}$

5 0

2 years ago