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Veseljchak [2.6K]
2 years ago
15

Which is the graph of f(x)= √x

Mathematics
1 answer:
4vir4ik [10]2 years ago
5 0

Answer:

ion know

Step-by-step explanation:

you figure it out

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12=5x-13x-44. answer it in multi-step equations
fomenos

12=5x-13x-44

combine like terms

12 = -8x -44

add 44 to each side

12+44 = -8x

56= -8x

divide by -8 on each side

56/-8 = x

-7 =x


5 0
3 years ago
Read 2 more answers
In a parallelogram what can you say about the consecutive angle
Sav [38]

Answer:

Sum of two consecutive angle is always 180°

Step-by-step explanation:

In a parallelogram what we can you say about the consecutive angle is that they are supplementary to each other. By supplementary we mean that the sum of the two angles is always 180°

4 0
3 years ago
What part of the circumference is an arc whose measure is 30°?
Radda [10]
Circumference of a circle:
C = 2 r π;
Length of an arc:
L = r π α / 180°
L = r π · 30° / 180° = r π /6
r π /6   :   2 r π = 1/6 : 2 = 1/12
Answer: A ) 1/12 
7 0
3 years ago
Evaluate this: -10x^0
FinnZ [79.3K]

Answer:

-10.

Step-by-step explanation:

x^0 = 1 so its -10 * 1

= -10.

4 0
3 years ago
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A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
2 years ago
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