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2 years ago

Find the solution set of this inequality. Enter your answer in interval notation using grouping symbols. -2x+5<3

Mathematics

2 answers:

Andrej [43]2 years ago

8 0

Answer:

( 1 , ∞ )

Step-by-step explanation:

Interval notation rounds up to infinite

Fed [463]2 years ago

5 0

(1,infinity),,,,,,,,,

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If the temperature is dropping at a rate of 2° per hour how many hours will it take to drop 15°

Tju [1.3M]

Answer:

7 1/2

Step-by-step explanation:

If it drops 2 per hour, and you need it to drop 15, divide 15 by 2 and you’ll get 7.5 or 7 1/2

5 0

2 years ago

Radioactive Waste The rate at which radioactive waste is entering the atmosphere at time t is decreasing and is given by Pe^-kt,

Dmitry_Shevchenko [17]

Answer:

$\displaystyle{\int^{\infty}_0 {50e^{-0.04t}}\, dt}=1250$

Step-by-step explanation:

Given:

$T =\displaystyle{\int^{\infty}_0 {Pe^{-kt}} \, dt}$

where,

T = total amount of waste

P = 50, the initial rate

k = 0.04

t = time

$T =\displaystyle{\int^{\infty}_0 {50e^{-0.04t}} \, dt}$

now we need to solve this integral!

$T =\displaystyle{50\int^{\infty}_0 {e^{-0.04t}} \, dt}$

$T = \left|50\left(\dfrac{e^{-0.04t}}{-0.04}\right)\right|^{\infty}_0$

$T = \left|-1250e^{-0.04t}\right|^{\infty}_0$

$T = (-1250e^{-0.04(\infty)})-(-1250e^{-0.04(0)})$

when any number has a power of negative infinity it is 0. because: $a^-{\infty} = \dfrac{1}{a^{\infty}} = \dfrac{1}{\infty} = 0$ , like something being divided by a very large number!

$T = (-1250(0))-(-1250e^0)$

$T = 1250$

this is the total amount of waste

6 0

2 years ago

Can you help me with these please 30 points

garik1379 [7]

Answer: 1. densities

2. i don’t have the passage but i assume it to be A or D

3. it will sink so A

Step-by-step explanation:

the density must be lower than 1 g/cm3 for it to float

3 0

3 years ago

Read 2 more answers

Match the numerical expressions to their simplest forms.

Aloiza [94]

Answer:

$(a^6b^1^2)^\frac{1}{3}$ = $a^2b^4$

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$ = $a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$ = $a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$ = $ab^3$

Step-by-step explanation:

Simplify each of the expressions:

$(a^6b^1^2)^\frac{1}{3}$

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

$(a^6b^1^2)^\frac{1}{3}$

$a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}$

Simplify,

$a^2b^4$

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$

Rewrite as square roots:

$\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}$

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

$\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}$

Simplify,

$\sqrt{a^5b^3}*\sqrt{ab}$

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

$\sqrt{a^5b^3*ab}$

Simplify,

$\sqrt{a^6b^4}$

Since this operation is taking the square root, divide the exponents in half to do this operation:

$a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

$(\frac{a^8}{b^-^4})^\frac{1}{4}$

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

$(a^8b^4)^\frac{1}{4}$

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

$\sqrt[4]{a^8b^4}$

SImplify, divide the exponents by (4) to simulate taking the quartic root,

$a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

$(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}$