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4 years ago

Solve -2(2x+5) -3(x-1)

Mathematics

1 answer:

OLEGan [10]4 years ago

8 0

<h2>Hello!</h2>

The answer is:

$-2(2x+5)-3(x-1)=-7x-7\\$

To solve the problem, we need to perform the operations and then, add like terms.

We need to apply the distributive property, which is defined by the following way:

$a(b+c)=ab+ac$

Also, we need to remember how to add like terms. Like terms are terms that share the same exponent and the same variable, for example:

$3x+5x+x^{2}=8x+x^{2}$

We were able to add only the first two terms since they are like terms, both are sharing the same exponent and the same variable.

So, we are given the expression:

$-2(2x+5)-3(x-1)$

Then, solving we have:

$-2(2x+5)-3(x-1)=-4x-10-(3x-3)\\\\-4x-10-(3x-3)=-4x-10-3x+3\\\\-4x-10-3x+3=-4x-3x-10+3\\\\-4x-3x-10+3=-7x-7$

Hence, we have that the answer is:

$-2(2x+5)-3(x-1)=-7x-7\\$

Have a nice day!

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Substance X is found in the bloodstream of humans. A random sample of 75 males had a mean content of 28 ppm of Substance X in th

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Answer:

H0: u1 - u2 = 0

H1: u1 - u2 ≠ 0

Step-by-step explanation:

This question requires us to state the hypothesis to be used for this test to know if males and females have the same amount of substance X. I have gone further to also calculate the test statistics too.

Null hypothesis:

H0: u1 - u2 = 0

Alternative hypothesis:

H1: u1 - u2 ≠ 0

X1 = 28, sd1 = 14.1, n1 = 75

X2 = 33, sd2 = 9.5, n2 = 50

We have a 2 tailed test

Alpha = 1-0.8812

= 0.1188

Critical value = 1.56

If z<-1.56 or z>1.56

Reject null hypothesis

Test statistic

= (28-33)/14.1²/75 + 9.5²/50

= -2.37

Therefore we reject the H0 because -2.37 is less than -1.56

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One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number o

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To be clear, the given relation between time and female population is an integral:
 $t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP$

The problem says that r = 1.2 and S = 400, therefore substituting:
 $t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]}
} \, dP$ 

= $
\int { \frac{P+400}{P(0.2P - 400)} } \, dP$

In order to evaluate this integral, we need to write this rational function in a simpler way:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} +
\frac{B}{(0.2P - 400)}$ 

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P -
400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$ 

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
 = P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
 $\left \{ {{0.2A + B = 1} \atop {-400A =
400}} \right.$ 

Which can be easily solved:
 $\left \{ {{B=1.2} \atop {A=-1}} \right.$ 

Therefore, our integral can be written as:
 $t = \int { \frac{P+400}{P(0.2P - 400)} } \,
dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$ 
= $- \int { \frac{1}{P} \, dP +
1.2\int { \frac{1}{0.2P-400} } \, dP$ 
= $- \int { \frac{1}{P} \, dP +
6\int { \frac{0.2}{0.2P-400} } \, dP$ 
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5) 
 Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)

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