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Vladimir79 [104]
3 years ago
11

Help me please!!!! I will mark Brainliest

Mathematics
1 answer:
ryzh [129]3 years ago
8 0

Answer:

32 pi

Step-by-step explanation:

im sorry if im wrong im not the smartest haha

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How do you graph f(x)= |x-2|+5
ElenaW [278]
For this we are going to use the equation f(x)=a|x+h|+k
Since there is no a shown, we know that a=1. This means the slope of the graph by 1's. (slope=1 one on the right side and slope=-1 on the left, because absolute value graphs are shaped like a "v".)
Since h=-2, just move the function to the left 2 units.
Since k=5, move the function up 5 units.
This basically means the vertex of the function is at the point (-2,5).

So put the point (-2,5) on your graph, then make the v-shape with a slope of 1 on the right side, and -1 on the left. 

Sometimes if you aren't sure how the graph will look, just plug x-values into the function to get their corresponding y-values. Then see how the graph looks to check.

5 0
3 years ago
Which function has a vertex at the orgin
amm1812
Explain a little more
5 0
4 years ago
If ΔPQR ≅ ΔJKL and ∠P = 35º, then which other angle also equals 35º. a ∠J b ∠K c ∠L d None of these choices are correct.
aivan3 [116]

Answer:

A: Angle J

Step-by-step explanation

If the two triangles are congruent then their corresponding angles are also congruent. This means that angle P is congruent to angle J, angle Q is congruent to angle K and angle R is congruent to angle L. Hope this helps :D

3 0
3 years ago
What is the answer to this problem
Gre4nikov [31]
Baxter ate 1/12, so now the box is 3/4 - 9/12 full. So, (1/12) + (9/12) = 10/12.

10/12 of full box was there before Baxter ate.
8 0
4 years ago
Let a=<-4,-3,5> Find a unit vector in the same direction as having positive first coordinate.
Rashid [163]

Answer:

The value is u =  < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>

Step-by-step explanation:

From the question we are told that

The vector is a=<-4,-3,5>

Generally the unit vector is u = ay

Here y represent the y-coordinate

So

u =y

=> u =

Generally the resultant of a unit vector is 1

So

|u| = \sqrt{ (-4y)^2 + (-3y)^2 + (5y)^2} = 1

Hence

|u| = \sqrt{ 16y^2 + 9y^2 + 25y^2} =  1

Taking the square of both sides

16y^2 + 9y^2 + 25y^2 =  1

=> 50y^2 =  1

=> y =  \pm \frac{1}{\sqrt{50}}

=> y =  \pm \frac{1}{5 \sqrt{2}}

Rationalizing

=> y =  \pm \frac{\sqrt{2}}{10}

Given that the first coordinate is positive

y = - \frac{\sqrt{2}}{10}

Hence the unit vector is

u =

=> u =  < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>

4 0
3 years ago
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