Use the law of exponets to rewrite (5x3)^7

Please can someone help me I'm confused (Brainliest to the person who gets it )

dimaraw [331]

Answer:

$90 = 60p

p = $1.50

Step-by-step explanation:

Brainliest Please!

8 0

3 years ago

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20 × 20 - 4 solve and show work

Paul [167]

Answer:

20×20-4

=400-4

=396

hope it helps u....

plz mark as brainliest...

5 0

3 years ago

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The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section

lisabon 2012 [21]

Answer:

$\therefore y_2(x)=-\frac{e^{-6x}}{8}$

The general solution is

$y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}$

Step-by-step explanation:

Given differential equation is

y''-4y'+4y=0

and $y_1(x)=e^{2x}$

To find the $y_2(x)$ we are applying the following formula,

$y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} \ dx$

The general form of equation is

y''+P(x)y'+Q(x)y=0

Comparing the general form of the differential equation to the given differential equation,

So, P(x)= - 4

$\therefore y_2(x)=e^{2x}\int \frac{e^{-\int 4dx}}{(e^{2x})^2}dx$

$=e^{2x}\int \frac{e^{-4x}}{e^{4x}}dx$

$=e^{2x}\int e^{-4x-4x} \ dx$

$=e^{2x}\int e^{-8x} \ dx$

$=e^{2x}. \frac{e^{-8x}}{-8}$

$=-\frac{e^{-6x}}{8}$

$\therefore y_2(x)=-\frac{e^{-6x}}{8}$

The general solution is

$y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}$

4 0

3 years ago

Plz help I need these done ASAP

Olegator [25]

Answer: a > 4

Step-by-step explanation:

-5 + a > -1

+5 +5

a > 4

Graph: 4 o-----------→

Interval Notation: (4, ∞)

7 0

3 years ago

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Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

earnstyle [38]

Answer:

$t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167$

$p_v =2*P(t_{23}$

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

Step-by-step explanation:

Data given

$n_1 =10$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =1085$ represent the sample mean for the group 1

$\bar X_2 =1034$ represent the sample mean for the group 2

$s_1=52$ represent the sample standard deviation for group 1

$s_2=61$ represent the sample standard deviation for group 2

We are assuming that the two samples are normally distributed with equal variances and that means:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

System of hypothesis

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

The statistic is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The degrees of freedom are given by:

$n_1+n_2 -2$

And the pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Replacing we got:

$S^2_p =\frac{(10-1)(52)^2 +(15 -1)(61)^2}{10 +15 -2}=3323.043$

And the deviation would be:

$S_p=57.646$

The degrees of freedom are:

$df=10+15-2=23$

The statistic would be:

$t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167$

The p value would be

$p_v =2*P(t_{23}$

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

6 0

3 years ago