All categories

3 years ago

The range of y=√x-5-1 is

Mathematics

2 answers:

kobusy [5.1K]3 years ago

8 0

Answer:

-1

Step-by-step explanation:

Lady_Fox [76]3 years ago

6 0

Answer:

[-1, ∞ )

Step-by-step explanation:

The use of parentheses is essential here. I am assuming that you meant:

y=√(x-5) - 1

Note that √(x-5) has the range [0, ∞ ). The domain is [5, ∞ ).

Thus, y=√(x-5) - 1 has the range [-1, ∞ )

You might be interested in

Viewing C

Trava [24]

Answer:

ccrccr

Step-by-step explanation:

drcgtgtfttcctt

6 0

2 years ago

) Solve this equation: − 10 + = 8 + 2 − 18

nikitadnepr [17]

Answer:

I think the question is wrong check ur question and I will answer

4 0

2 years ago

A line passes through (−1, 7) and (2, 10).

Murljashka [212]

The formula is y=mx+b
To get the slope or m, use this formula
(the second y minus the first y)/(the second x minus the first x)
Now set it up.
(10-7)/(2--1)
3/3=slope is 1.
y=1x+b

Insert one of the points for x and y.
i will do (-1,7)
7=1(-1)+b
7=-1+b
8=b
Insert this into the final equation:
y=1x+8

Try it out. If you're not sure, try both points. If it works, then you set it up correctly.

6 0

3 years ago

Read 2 more answers

Can anybody help out this math question it’s about graph please

malfutka [58]

Neither one of the slopes are going to be minus. It means you are travelling backwards in time, which is wonderful if you are a sci-fi fan, but not so good if you are Sharon. A and D has Sharon going from 70 to 0. That can't be happening so both are wrong.

Now you have to decide between B and C. The intersection point has Sharon going upwards until she is 20. She started out at 70. The graph has John starting at 70. That's not right.

So we've eliminated A,D and now C.

The answer must be B. They meet when Sharon is 90 and John is about 22.5 which is what it should be.

6 0

2 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

3 years ago

The range of y=√x-5-1 is​

The range of y=√x-5-1 is