Please help I'm begging this is really hard

4(4m-3)-m(m-5)=-52 need all the steps

nadezda [96]

$\huge\mathcal {♨Answer♥}$

$\large\texttt{Simplifying: }$

4(4m -3) + -1m(m + -5) = -52

$\large\texttt{Reorder the terms: }$

4(-3 + 4m) + -1(m + -5) = -52

(-3 * 4 + 4m * 4) + -1(m + -5) = -52

(-12 + 16m) + -1(m + -5) = -52

$\large\texttt{Reorder the terms: }$

-12 + 16m + -1(-5 + m) = -52

-12 + 16m + (-5 * -1 + m * -1) = -52

-12 + 16m + (5 + -1m) = -52

$\large\texttt{Reorder the terms: }$

-12 + 5 + 16m + -1m = -52

$\large\texttt{Combine like terms: }$

-12 + 5 = -7

-7 + 16m + -1m = -52

$\large\texttt{Combine like terms: }$

16m + -1m = 15m

-7 + 15m = -52

$\large\texttt{Solving: }$

-7 + 15m = -52

-7 + 7 + 15m = -52 + 7

$\large\texttt{Combine like terms: }$

-7 + 7 = 0

0 + 15m = -52 + 7

15m = -52 + 7

$\large\texttt{Combine like terms: }$

-52 + 7 = -45

15m = -45

$\large\texttt{Divide each side by '15'. }$

15m ÷ 15 = -45 ÷ 15

m = -3

$\large\texttt{Simplifying: }$

m = -3

☆...hope this helps...☆

_♡_mashi_♡_

8 0

2 years ago

Evaluate: 2a + 4b when a = 10 & b = 6

Tcecarenko [31]

Answer: Option C: 44

Step-by-step explanation:

so, here we have the equation:

H(a,b) = 2a + 4b

"evaluate" means change the values of the variables for specific values, here we must replace the "a" for 10, and the "b" for a 6.

So we have:

H(10, 6) = 2*10 + 4*6 = 20 + 24 = 44

7 0

3 years ago

Whats 2+2*89*546/5^2+768/678*100000

Natasha_Volkova [10]

330987894/2825 you can check your answers by using a calculator such as online cal.

7 0

3 years ago

Find the area of this figure 28 cm 23.7 cm 8 cm 5.7 cm

maks197457 [2]

Yes you do it ima was a great time and ya I wanna ya got to be like that you were just playing with him lol is that he cute too lol

3 0

3 years ago

Read 2 more answers

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .

5 0

4 years ago