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2 years ago

9. Square Root Functions

Mathematics

1 answer:

topjm [15]2 years ago

8 0

Answer:

y = √(x - 3)

Step-by-step explanation:

The graph shown has exactly the same shape as does the graph of y = √x, EXCEPT that the entire graph of y = √x has been translated 3 units to the right. Thus, the function describing this graph is y = √(x - 3).

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What is the surface area of the cube shown?<br> square centimeters

Mnenie [13.5K]

Answer: 253.5

Step-by-step explanation:

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2 years ago

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the figure below shows a circle centre of radius 10 cm the chord PQ=16cm calculate the area of the shaded region

m_a_m_a [10]

Step-by-step explanation:

OPQ originally forms a sector, the formula for sector is

$\frac{x}{360} \pi {r}^{2}$

where x is the degrees of rotation between the two radii.

We know three sides length and is trying to find an angle between the radii so we can use law of cosines which states that

$16 = \sqrt{10 {}^{2} + 10 {}^{2} - 2(100) \times \cos(o) }$

This isn't the standard formula, it's for this problem

$16 = \sqrt{200 - 200 \times \cos(o) }$

$16 = \sqrt{200 - 200 \cos(o) }$

$256 = 200 - 200 \cos(o)$

$56 = - 200 \cos(o)$

$- \frac{7}{25} = \cos(o)$

$\cos {}^{ - 1} ( - \frac{7}{25} ) = \cos {}^{ - 1} ( \cos(o) )$

$106.26 = o$

So we found our angle of rotation, which is 106.26

Now, we do the sector formula.

$\frac{106.26}{360} (100)\pi$

$\frac{10626}{360} \pi$

is the area of sector.

Now let find the area of triangle, we can use Heron formula,

The area of a triangle is

$\sqrt{s(s - a)(s - b)(s - c)}$

where s is the semi-perimeter.

To find s, add all the side lengths up, 10,10,16 and divide it by 2.

Which is

$\frac{36}{2} = 18$

$\sqrt{18(18 - 10)(18 - 10)(18 - 16)}$

$\sqrt{18(8)(8)(2)}$

$\sqrt{(36)(64)}$

$6 \times 8 = 48$

So our area of the triangle is 48. Now, to find the shaded area subtract the main area,( the sector of the circle) by the area of the triangle so we get

$\frac{10626}{360} - 48$

Which is an approximate or

44.73

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2 years ago

What is the answer to 3X - 5=16

sineoko [7]

The answer is x=7. I hope this helps. Tell me if you need an explanation!

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3 years ago

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Line Segment DE is parallel to side BC of right triangle ABC. CD = 3, DE = 6, and EB = 4. Compute the area of quadrilateral BCDE

dybincka [34]

The area of quadrilateral BCDE = 20.4 sq. units

Let AD = x and AE = y.

Since ΔABC and ΔAED are similar right angled triangles, we have that

AC/AD = AB/AE

AC = AD + CD

= x + 3.

Also, AB = AE + EB

= y + 4

So, AC/AD = AB/AE

(x + 3)/x = (y + 4)/y

Cross-multiplying, we have

y(x + 3) = x(y + 4)

Expanding the brackets, we have

xy + 3y = xy + 4x

3y = 4x

y = 4x/3

In ΔAED, AD² + AE² = DE².

So, x² + y² = 6²

Substituting y = 4x/3 into the equation, we have

x² + y² = 6²

x² + (4x/3)² = 6²

x² + 16x²/9 = 36

(9x² + 16x²)/9 = 36

25x²/9 = 36

Multiplying both sides by 9/25, we have

x² = 36 × 9/25

Taking square root of both sides, we have

x = √(36 × 9/25)

x = 6 × 3/5

x = 18/5

x = 3.6

Since y = 4x/3,

Substituting x into the equation, we have

y = 4 × 3.6/3

y = 4.8

To find the area of quadrilateral BCDE, we subtract the area of ΔAED from area of ΔABC.

So, area of quadrilateral BCDE = area of ΔABC - area of ΔAED

area of ΔABC = 1/2 AC × AB

= 1/2 (x + 3)(y + 4)

= 1/2(3.6 + 3)(4.8 + 4)

= 1/2 × (6.6)(8.8)

= 1/2 × 58.08

= 29.04 square units

area of ΔAED = 1/2 AD × AE

= 1/2xy

= 1/2 × 3.6 × 4.8

= 1/2 × 17.28

= 8.64 square units

area of quadrilateral BCDE = area of ΔABC - area of ΔAED

area of quadrilateral BCDE = 29.04 sq units - 8.64 sq units

area of quadrilateral BCDE = 20.4 sq. units

So, the area of quadrilateral BCDE = 20.4 sq. units

Learn more about area of a quadrilateral here:

brainly.com/question/19678935

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2 years ago

Which of the following would be the best strategy?

Dvinal [7]

Answer:

I think the answer is D

Step-by-step explanation:

I hope it was correct...

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3 years ago