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2 years ago

In the figure diameter DE is drawn with the midpoint of the circle at C(-3; 2).

Mathematics

1 answer:

Lelechka [254]2 years ago

7 0

Answer:

Step-by-step explanation:

$gradient = \frac{y2 - y1}{x2 - x1}$

$y2 = - 3 \\ y1 = 3 \\ x2 = 2 \\ x1 = 3$

$\frac{ - 3 - 1}{2 - 3} = \frac{ - 4}{ - 1} = 4$

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Ian wants to run 412 miles to get ready for his upcoming race.

tankabanditka [31]

Answer:

137.3 miles apart from each other

Step-by-step explanation:

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3 years ago

Read 2 more answers

H(x) = x^2+6 g(x) = 8x-5 <br><br> show that hg(x)=19 simplifies to 16x^2 - 20x + 3 = 0

Studentka2010 [4]

Answer:

h(x)=x²+6 and g(x)=8x-5

prove that hg(x)=19

Step-by-step explanation:

step 1: hog(x)=h[g(x)]

step 2: h(8x-5)

step 3: h(8x-5)= (8x-5)²+6

step 4: (8x-5)(8x-5)+6

step 5: 16x² +15x +6

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3 years ago

What is the remainder when 4x^3+2x^2-18+38/x-3

Trava [24]

Answer:

The remainder is 2

Step-by-step explanation:

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7 0

4 years ago

Find the domain of the function f(x)=<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E3-16x%7D" id="TexFormula1" title="\sqrt{x^3

Anon25 [30]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the function

$f\left(x\right)=\sqrt{x^3-16x}$

We know that the domain of the function is the set of input or arguments for which the function is real and defined.

In other words,

Domain refers to all the possible sets of input values on the x-axis.

Now, determine non-negative values for radicals so that we can sort out the domain values for which the function can be defined.

$x^3-16x\ge 0$

as x³ - 16x ≥ 0

$\left(x+4\right)\left(x-4\right)\ge \:0$

Thus, identifying the intervals:

$-4\le \:x\le \:0\quad \mathrm{or}\quad \:x\ge \:4$

Thus,

The domain of the function f(x) is:

$x\left(x+4\right)\left(x-4\right)\ge \:0\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-4\le \:x\le \:0\quad \mathrm{or}\quad \:x\ge \:4\:\\ \:\mathrm{Interval\:Notation:}&\:\left[-4,\:0\right]\cup \:[4,\:\infty \:)\end{bmatrix}$

And the Least Value of the domain is -4.

3 0

3 years ago

Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AB = 8

Delicious77 [7]

Answer:

x = 32

Step-by-step explanation:

∠BCA = ∠DBA (90 - ∠DBC)

∠A = ∠A

ΔABD similar to ΔACB

AC/AB = AB/AD

x / 8 = 8 / 2

x = 32

4 0

3 years ago