All categories

4 years ago

Which point has the coordinates (-2.5, 5.5)?

Mathematics

2 answers:

frosja888 [35]4 years ago

7 0

The answer is A point E

BaLLatris [955]4 years ago

4 0

Answer:

it would be H and F because if u line up the number with the cordnets you will see that the number match with it.

hope this helps. have a nice day..

You might be interested in

A right triangle ABC is shown below:

torisob [31]

Answer:

3 units wide

Step-by-step explanation:

answer is 3 units wide

6 0

3 years ago

Rectangle A'B'C'D' is the image of rectangle ABCD after a dilation centered at point P.

n200080 [17]

What is the question? In other words, what does the question want us to find?

3 0

3 years ago

Eliminate the parameter:<br> x = 2cost-1 y = 3sint+ 4

adell [148]

Answer:

Hi,

Step-by-step explanation:

$\left\{\begin {array} {ccc}x&=&2*cos(t)-1\\y&=&3*sin(t)+4\\\end{array}\right.\\\\\\\left\{\begin {array} {ccc}\dfrac{x+1}{2}&=&cos(t\\\dfrac{y-4}{3}&=&sin(t)\\\end{array}\right.\\\\Squaring\\\\\boxed{(\dfrac{x+1}{2})^2+(\dfrac{y-4}{3})^2&=1}\\$

The curve is an ellipse having (-1,4) as center and half-axis (2 and 6)

8 0

2 years ago

A squirrel feeder holds 4 1/2 cups of seeds. Another squirrel feeder

taurus [48]

Answer:

Step-by-step explanation:

5 0

3 years ago

In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.

DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

$(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2$

$+ 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5$

$5C_0(\frac{1}{ax})^5(2ax^2)^0 =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1 = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\$

$5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}$

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

6 0

3 years ago