All categories

2 years ago

Which of the following expanded forms represents the number words one thousand, six hundred eighty-nine?

Mathematics

1 answer:

Arisa [49]2 years ago

5 0

The answer is 1000+600+80+9. please give brainliest

You might be interested in

What are the coordinates of the vertices of the image?

Lady bird [3.3K]

Step-by-step explanation:

(2,4)===(4,-2)

(4,3)===(3,-4)

(2,3)===(3,-2)

8 0

2 years ago

Find the Diameter of the Circle with the following equation. Round to the nearest tenth.

Radda [10]

The diameter of the circle is 11.8 units

<h3>How to determine the diameter of the circle?</h3>

The circle equation is given as:

(x - 2)^2 + (y + 6)^2 = 35

A circle equation is represented as:

(x - a)^2 + (y - b)^2 = r^2

Where

Diameter = 2r

By comparing both equations, we have

r^2 = 35

Take the square root of both sides

r = 5.9

Multiply by 2

2r = 11.8

Hence, the diameter of the circle is 11.8 units

Read more about circle equation at:

brainly.com/question/10618691

#SPJ1

3 0

2 years ago

Which of the following functions is graphed below?

aleksklad [387]

Answer:

Step-by-step explanation:

the correct answer is A. y=|x-2|+3.

For more details see the attachment.

6 0

2 years ago

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

6 0

3 years ago

Find X using the image.

Temka [501]

9514 1404 393

Answer:

x = 2

Step-by-step explanation:

The square of the tangent is equal to the product of the secant lengths to the near and far circle intersections.

(x +4)² = (2x -1)(2x -1 +9)

0 = (2x -1)(2x +8) -(x +4)² = (4x² +14x -8) -(x² +8x +16)

0 = 3x² +6x -24 = 3(x² +2x -8) = 3(x +4)(x -2)

This has solutions x = -4 and x = 2. Only the latter works in this geometry.

x = 2

7 0

3 years ago