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Alinara [238K]
3 years ago
9

Describe the location of the point having the following coordinates. negative abscissa, zero ordinate between Quadrant II and Qu

adrant IV Quadrant II between Quadrant I and Quadrant IV between Quadrant II and Quadrant III

Mathematics
1 answer:
Marianna [84]3 years ago
6 0

Answer:

The location of the point is between Quadrant II and Quadrant III

Step-by-step explanation:

we know that

The abscissa refers to the x-axis  and ordinate refers to the y-axis

so

in this problem we have

the coordinates of the point are (-x,0)

see the attached figure to better understand the problem

The location of the point is between Quadrant II and Quadrant III

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A=34.5. B=29.04. C= 204.7 D= 18.83
ahrayia [7]

The answer is B.) 29.04

you find c and then add it to a and b

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3 years ago
D is the midpoint of line segment GH. GH = 3x - 1 and DH = 16. Find x.​
docker41 [41]

Answer:

3x-1=16*2

3x=32+1

3x=33/:3

x=11

Step-by-step explanation:

3 0
4 years ago
The product of x and the sum of 6 and 8 times the square of x
Vaselesa [24]

Answer:

x(6 + 8x²) or 6x + 8x³.

Step-by-step explanation:

"The square of x" can be represented by x² and 8 times that would be 8 * x² or 8x². The sum of 6 and 8x² can be represented by 6 + x² and the product of x and 6 + x² can be represented by x * (6 + 8x²) or x(6 + 8x²) which simplifies to 6x + 8x³.

7 0
3 years ago
Read 2 more answers
For 0 ≤ θ < 2 π what are the solutions to sin^2(θ) =2sin^2(θ/2)
gregori [183]

Recall the half-angle identity for sine,

sin²(x/2) = (1 - cos(x))/2

Then the given equation is identical to

sin²(θ) = 1 - cos(θ)

Also recall the Pythagorean identity,

sin²(θ) + cos²(θ) = 1

Then we rewrite the equation as

1 - cos²(θ) = 1 - cos(θ)

Factoring the left side, we have

(1 - cos(θ)) (1 + cos(θ)) = 1 - cos(θ)

and so

(1 - cos(θ)) (1 + cos(θ)) - (1 - cos(θ)) = 0

and we factor this further as

(1 - cos(θ)) (1 + cos(θ) - 1) = 0

which gives

cos(θ) (1 - cos(θ)) = 0

Then either

cos(θ) = 0   or   1 - cos(θ) = 0

cos(θ) = 0   or   cos(θ) = 1

[θ = arccos(0) + 2nπ   or   θ = -arccos(0) + 2nπ]

…   or   [θ = arccos(1) + 2nπ   or   θ = -arccos(1) + 2nπ]

(where n is any integer)

[θ = π/2 + 2nπ   or   θ = -π/2 + 2nπ]   or   [θ = 0 + 2nπ]

In the interval 0 ≤ θ < 2π, we get three solutions:

• first solution set with n = 0   ⇒   θ = π/2

• second solution set with n = 1   ⇒   θ = 3π/2

• third solution set with n = 0   ⇒   θ = 0

So, the first choice is correct.

6 0
3 years ago
Addison's Cafe offers two kinds of espresso: single-shot and double-shot. Yesterday afternoon, the cafe sold 70 espressos in all
Valentin [98]

Answer:

The number of single-shot espressos sold by the cafe was 49.

Step-by-step explanation:

It is provided that Addison's Cafe offers two kinds of espresso: single-shot and double-shot.

Total number of espressos sold by the cafe yesterday afternoon was,

<em>N</em> = 70

The proportion of single-shot espresso sold was, <em>p</em> = 0.70.

Let <em>X</em> = number of single-shot espressos sold.

Compute the number of single-shot espressos sold by the cafe as follows:

E(X)=N\times p

         =70\times 0.70\\=49

Thus, the number of single-shot espressos sold by the cafe was 49.

6 0
3 years ago
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