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3 years ago

(-9q^3+-8q) + (3q^2-q-6q^3) I need to find the sum

Mathematics

2 answers:

RUDIKE [14]3 years ago

8 0

Answer:

−15q3+3q2−9q

Step-by-step explanation:

−9q3−8q+3q2−q−6q3

=−9q3+−8q+3q2+−q+−6q3

Combine Like Terms:

=−9q3+−8q+3q2+−q+−6q3

=(−9q3+−6q3)+(3q2)+(−8q+−q)

=−15q3+3q2+−9q

xxTIMURxx [149]3 years ago

7 0

Answer:

-3q(4q^2+q+3)

Step-by-step explanation:

To solve this question we will have to open the bracket first

So let's solve

(-9q^3+(-8q)+(3q^2-q-6q^3)

Open the bracket first

(-9q^3-8q)+3q^2-q-6q^3

-9q^3-8q+3q^2-q-6q^3

Let's collect like terms

-9q^3-6q^3+3q^2-8q-q

-12q^3+3q^2-9q

We can actually factorise

-3q(4q^2+q+3)

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The radical form plz someone help

lapo4ka [179]

Answer:

it cannot be converted to radical form

Step-by-step explanation:

5 0

3 years ago

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of

saw5 [17]

Let A(t) denote the amount of salt in the tank at time t.

Salt flows in at a rate of

(1 lb/gal) * (3 gal/min) = 3 lb/min

and flows out at a rate of

(A(t)/(200 + t) lb/gal) * (2 gal/min) = 2 A(t)/(500 + t)

(in case you're unsure about the denominator: the tank starts off with 200 gal of solution, and each minute solution flows in at a rate of 3 gal/min and thus the tank gains (3 gal/min) * (1 min) = 3 gal. At the same time, solution flows out at a rate of 2 gal/min and thus the tank loses 2 gal, giving a net change in volume of (3 - 2)*t = t gal)

Then the net rate of salt flow is given by the ODE,

$\dfrac{\mathrm dA(t)}{\mathrm dt}-\dfrac{2A(t)}{200+t}=3$

Multiply both sides by $(200+t)^{-2}$ :

$(200+t)^{-1}\dfrac{\mathrm dA(t)}{\mathrm dt}-2(200+t)^{-3}A()=3(200+t)^{-2}$

$\implies\dfrac{\mathrm d}{\mathrm dt}\bigg((200+t)^{-2}A(t)\bigg)=3(200+t)^{-2}$

Integrating both sides and solving for $A(t)$ gives

$(200+t)^{-2}A(t)=-\dfrac3{200+t}+C$

$A(t)=-2(200+t)+C(200+t)^2$

The tank starts off with 100 lb of salt in solution, so $A(0)=100$ and we find

$100=-2(200)+C(200)^2\implies C=\dfrac1{80}$

and so

$A(t)=-2(200+t)+\dfrac{(200+t)^2}{80}=\dfrac{(200+t)(40+t)}{80}$

The tank will begin to overflow once the volume of solution reaches 500 gal; this happens when

$500=200+t\implies t=300$

or 300 minutes or 5 hours after solution starts flowing. At this point, the tank will contain

$A(300)=2125$

or 2125 lb of salt.

Theoretically, the amount of salt in the tank will increase forever, since $A(t)\to\infty$ as $t\to\infty$ .

6 0

4 years ago

Variances indicate a.the cause of the variance. b.who is responsible for the variance. c.when the variance should be investigate

Nezavi [6.7K]

Option c. is the correct answer.

When the variance should be investigated.

A variance is the difference between a result's actual value and what was anticipated or budgeted. Variances can be either positive or negative. When the actual is higher than the projected or budgeted amount, a favorable variance occurs. The variance is unfavorable when actuals fall short of the amounts allocated in the budget. A variance therefore shows whether or not the actual performance is proceeding as expected.

While,

Option a. is inaccurate. As, the cause of the variance can vary, depending on factors like changes in delivery costs, market costs, production techniques, the use of subpar materials, seasonal factors, or even a simple mistake. Therefore, additional research and subsequent actions are needed to identify the cause of the variance. The cause of a variance cannot be identified by the variance itself.

Option b. is inaccurate. A variance only shows the discrepancy between the anticipated and actual amounts. It makes no mention of who is to blame for the deviation.

Option d. is inaccurate. After variances are identified, one of the most important steps is to investigate them. When actual performances significantly deviate from anticipated outcomes, variations are typically investigated.

Find more on Variance at : brainly.com/question/13287252

#SPJ4

3 0

2 years ago

g A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50. In a rando

Tema [17]

Answer:

We accept the null hypothesis, that is, the the mean amount of garbage per bin is not different from 50.

Step-by-step explanation:

A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50.

This means that the null hypothesis is:

$H_{0}: \mu = 50$

And the alternate hypothesis is:

$H_{a}: \mu \neq 50$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

50 tested at the null hypothesis:

This means that $\mu = 50$

In a random sample of 36 bins, the sample mean amount was 50.67 pounds and the sample standard deviation was 3.9 pounds.

This means that $n = 36, X = 50.67, \sigma = 3.9$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{50.67 - 50}{\frac{3.9}{\sqrt{36}}}$

$z = 1.03$

p-value:

Since we are testing if the mean is differente from a value and the z-score is positive, the pvalue is 2 multipled by 1 subtracted by the pvalue of z = 1.03.

z = 1.03 has a pvalue of 0.8485

1 - 0.8485 = 0.1515

2*0.1515 = 0.3030

0.3030 > 0.01, which means that we accept the null hypothesis, that is, the the mean amount of garbage per bin is not different from 50.

5 0

3 years ago

What is the product?

Bond [772]

Answer:

it is letter d or the last answer

Step-by-step explanation:

i just multiplied the 4 and -3, 3 and 1, -1 and -2, -1 and 2

6 0

3 years ago