Enter the range of values for x: B 15 2x - 10 C 48° A 18 5

How to graph -x+y=-6 , 3x+5y=2

Grace [21]

Solve for y for both

6 0

3 years ago

Look at the table to answer the question.

Alexus [3.1K]

Answer:

Your power function is y=-2 $x^{2}$

Step-by-step explanation:

The way I found it was.

0=0, so, the function multiply by zero, and have no other term to add

A number (one in a case) gives as a result: -2. So, one, elevated to ANY power results in one, every time. So, I have 1, as a factor and -2 as a result. One is as well a factor that delivers a result equal to the other factor. (3)(1)=3 , (8987)(1)=8987. So, the other factor must be -2

Then I checked all the table, and the results were consistent.

5 0

3 years ago

Translate the phrase into an algebraic expression the product of c and 8

MAVERICK [17]

Product is multiplying. when you multiply a letter with a number you don't put any symbols so it would be 8c

6 0

3 years ago

Find the volume of the composite space figure to the nearest whole number.

Romashka [77]

Answer:

d. 944 mm^3

Step-by-step explanation:

The area of a circle is given by ...

A = πr² . . . . . where r is the radius, half the diameter

The area of a circle with diameter 9 mm is ...

A = π(4.5 mm)² = 20.25π mm²

The area of the semi-circular end of the prism is half this value, or ...

semicircle area = (1/2)(20.25π mm²) = 10.125π mm² ≈ 31.809 mm²

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The area of the rectangular portion of the end of the prism is the product of its width and height:

A = wh = (9 mm)(6 mm) = 54 mm²

Then the base area of the prism is ...

base area = rectangle area + semicircle area

= (54 mm²) +(31.809 mm²) = 85.809 mm²

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This base area multiplied by the 11 mm length of the prism gives its volume:

V = Bh = (85.809 mm²)(11 mm) ≈ 944 mm³

The volume of the composite figure is about 944 mm³.

7 0

3 years ago

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0

3 years ago