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3 years ago

Solve 2y + 12 < 42 ???

Mathematics

1 answer:

Gemiola [76]3 years ago

8 0

Answer:

y<15

Step-by-step explanation:

2y+12=42

-12 -12

2y=30 divide each side by 2

y<15

(i wrote the first answer)

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10.Adrienne is going to deposit $8,500 in an account that earns 2.5% interest for 360 months. How much more interest will she ea

suter [353]

Answer:

Difference= $3,090.15 in favor of compounded interest

Step-by-step explanation:

Giving the following information:

Present value (PV)= $8,500

Ineterest (i)= 0.025/12= 0.00208

Number of periods (n)= 360 months

We will calculate the future value of each option and determine the difference:

Simple interest:

FV= (PV*i*n) + PV

FV= (8,500*0.00208*360) + 8,500

FV= $14,864.8

Compounded interest:

FV= PV*(1+i)^n

FV= 8,500*(1.00208^360)

FV= $17,958.95

Difference= $3,090.15

6 0

3 years ago

Brian has DVDs and CDs. The number of CDs he has can be modelled with the formula C = 2D + 11, where C represents the number of

wel

Answer: 15 DVDs

Step-by-step explanation:

C = 2D + 11

41 = 2D + 11

2D = 30

D = 15 DVDs

7 0

2 years ago

Is (-1,4) a solution of the inequality y<2x+5?

Sveta_85 [38]

Let's plug x= -1, y= 4 in the inequation, we have:
4< 2*(-1)+5
⇒ 4< -2+5
⇒ 4< 3 (false)

Therefore, (-1,4) is not a solution of the inequation y<2x+5~

6 0

3 years ago

In how many ways can 10 people form couples of two?

maxonik [38]

Answer:

It should be 10 raised to power 2 which is a hundred.

4 0

3 years ago

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