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2 years ago

Which net represents the figure you can just label it 1or 4

Mathematics

1 answer:

geniusboy [140]2 years ago

8 0

not sure second one is the answer

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Consider this sphere side the inder

Serhud [2]

Answer:a,c,e

Step-by-step explanation:

5 0

2 years ago

You can model that you expect a 1.25% raise each year that you work for a certain company. If you currently make $40,000, how ma

gizmo_the_mogwai [7]

Answer:

94 years

Step-by-step explanation:

We can approach the solution using the compound interest equation

$A= P(1+r)^t$

Given data

P= $40,000

A= $120,000

r= 1.25%= 1.25/100= 0.0125

substituting and solving for t we have

$120000= 40000(1+0.0125)^t \\\120000= 40000(1.0125)^t$

dividing both sides by 40,000 we have

$(1.0125)^t=\frac{120000}{40000} \\\\(1.0125)^t=3\\\ t Log(1.0125)= log(3)\\\ t*0.005= 0.47$

dividing both sides by 0.005 we have

$t= 0.47/0.005\\t= 94$

8 0

3 years ago

Lines b and c are parallel what is the measure of angle 6?

elena-s [515]

Answer:

54 degrees

Step-by-step explanation:

If these two lines are parallel, than angles 1 and 7 are supplementary.

13x+9+5x+9=180

18x+18=180

x+1=10

x=9

Angle 6 and 7 are vertical angles, so they are equal. Therefore, angle 6 is:

5x+9=5(9)+9=54

Hope this helps!!

3 0

3 years ago

What is the solution of the inequality Y+1<6

umka2103 [35]

Answer:

y < 5

Step-by-step explanation:

Given

y + 1 < 6 ( isolate y on the left side by subtracting 1 from both sides )

y < 5

6 0

3 years ago

Read 2 more answers

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