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3 years ago

HELP PLEASEE:(((((( THANk YOU

Mathematics

1 answer:

alina1380 [7]3 years ago

7 0

Answer:

(C). g(x) = ( $\frac{1}{4}$ x )²

Step-by-step explanation:

The correct answer is (C). g(x) = (  $\frac{1}{4}$  x )²

You might be interested in

(4 x 10) + (2 x 104) Give your answer in standard form

Goshia [24]

Answer: =248x

Step-by-step explanation: Hope this help :D

Let's simplify step-by-step.

4x(10)+2x(104)

=40x+208x

Combine Like Terms:

=40x+208x

=(40x+208x)

=248x

7 0

3 years ago

The accompanying table describes results from groups of 8 births from 8 different sets of parents. The random variable x represe

insens350 [35]

Using the discrete probability outputs given in the attached table ; the probability values of having exactly 6 ; and having 6 or more girls are :

P(X = 6) = 0.111

P(X ≥ 6) = 0.1

The probability of having exactly 6 girls can be defined as :

P(X = 6) = 0.111 (from the discrete probability table)

2.)

The probability of having 6 or more girls can be defined as :

P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8)

From the table attached :

P(X ≥ 6) = 0.111 + 0.014 + 0.003 = 0.1

Therefore, the probability of having exactly 6 girls is 0.111 while the probability of having 6 or more girls is 0.1

Learn more : brainly.com/question/18153040

5 0

3 years ago

X solve for the following inequality-1/2 p less than -16 which graph shows the correct solution

valkas [14]

Answer:

See attachment

Step-by-step explanation:

The given inequality is

$- \frac{1}{2}p \: < \: - 16$

We multiply both sides by -2 and reverse the inequality sign to get:

$- 2 \times - \frac{1}{2} p \: > \: - 2 \: \times - 16$

$p \: > \: 32$

We draw an open circle at 32 and draw an arrow to the right as shown in attachment.

8 0

3 years ago

Read 2 more answers

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

Joseph is building a cone using modeling clay. The cone has a radius of 6 cm and a height of 12 cm. Joseph adds additional clay,

dmitriy555 [2]

Cone Volume = (π • r² • h) ÷ 3

Original cone = (3.14 * 6^2 * 12) / 3

Original cone = 452.16 cc

Larger cone = (3.14 * 6^2 * 18) / 3 = 678.24 cc

Difference = (678.24 -452.16) = 226.08 cc

3 0

3 years ago

Read 2 more answers