All categories

2 years ago

HELP ME PLEASEEEEEEEEEEEE

Mathematics

1 answer:

olga nikolaevna [1]2 years ago

5 0

Answer:

Step-by-step explanation:

The first number is always x-coordinate. And amongst all only J and L has positive x-coordinates. With J y-coordinate is also positive so it should be L

You might be interested in

Can someone help me find the area

goldfiish [28.3K]

To find the volume of a container, you simply multiply: L(ength) x B(ase) x H(eight).

In total, you should get a volume of 1078in^3 (the bigger container) plus 175in^3 (the smaller container on top of the bigger one). The answer is 1,253in^3.

Hope this helped! :)

7 0

2 years ago

What number is 7 units to the left of -1? a.6 b.-6 c.-8 d.8

drek231 [11]

1 unit to the left of -1 on the number line is -2.

Thus, 7 units to the left of -1 is -8.

The appropriate choice is ...

... c. -8

6 0

2 years ago

Read 2 more answers

X ⋅ x = ? (1 point) jjjj

liraira [26]

Answer:

x-x = 0

x+x= 2x

x×x= x^2

have a nice day...

4 0

2 years ago

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

Then, the minimum point of this function is (3,1) and it corresponds to (A)

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

Then, the minimum point of this function is (0,1) and it corresponds to (C)

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

Then, the minimum point of this function is (2,3) and it corresponds to (B)

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

Then, the minimum point of this function is (2,-2) and it corresponds to (D)

I hope it helps you!

8 0

2 years ago

Can someone help me with this?

disa [49]

Alright, the answer is B and I’ll tell you why. Look at the problem and see the order that they write the triangles in. ABC and XYZ, this is telling us that the angles in order are congruent to each other. Hope this helps :)

7 0

3 years ago