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3 years ago

Please help me ASAP: Answer with : includes / not includes

Mathematics

1 answer:

Elodia [21]3 years ago

3 0

Answer:

Step-by-step explanation:

The interval [4, ∞) all real numbers 4 and greater, whereas the interval (4,∞) not includes the endpoint 4.

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Write two polynomials whose sum is x to the second power and whose difference is 1

Alex777 [14]

Answer:

Therefore, the first polynomial is:

$Z=\frac{1}{2}x^2+\frac{1}{2}\\$

the second polynomial is:

$Y=\frac{1}{2}x^2-\frac{1}{2}\\$

Step-by-step explanation:

We assume that: Z is first polynomial and Y is second polynomial.

Therefore, we have

$Z+Y=x^2\\\\Z-Y=1\\\\\implies Z+Y+Z-Y=x^2+1\\\\2Z=x^2+1\\\\Z=\frac{1}{2}x^2+\frac{1}{2}\\$

now, we get

$\implies Z+Y-Z+Y=x^2-1\\\\2Y=x^2-1\\\\Y=\frac{1}{2}x^2 -\frac{1}{2}$

Therefore, the first polynomial is:

$Z=\frac{1}{2}x^2+\frac{1}{2}\\$

the second polynomial is:

$Y=\frac{1}{2}x^2-\frac{1}{2}\\$

6 0

3 years ago

The distance traveled by a toy car can be modeled by D(t) = t3 − 3t2 + t − 3, where t > 3 represents time in seconds. If the

Bogdan [553]

Answer:

$t^2+1$

Step-by-step explanation:

We are given that

Distance traveled by a toy car

$D(t)=t^3-3t^2+t-3$

Where t>3

Time taken by car =t-3 seconds

We have to find the expression which represents the speed of the toy car.

We know that

$Speed=\frac{Distance}{time}$

Using the formula

$Speed=\frac{(t^3-3t^2+t-3)}{t-3}$

$Speed=\frac{t^2(t-3)+(t-3)}{t-3}$

$Speed=\frac{(t-3)(t^2+1)}{t-3}$

$Speed=t^2+1$

Hence, the expression which represents the speed of the toy car

$t^2+1$

5 0

3 years ago

Which mode of transportation did you choose? Why?

faust18 [17]

Answer:

land

Step-by-step explanation:

because,bus are more comfortable and easier other transportation are dangerous than land transportation ❤️

hope i helped u

8 0

3 years ago

CosA + cosB - cosC = -1 + 4cosA/2 cosB/2 sinC/2

Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

(cos A+ cos B)-cos C

$=2cos \frac{A+B}{2}cos \frac{A-B}{2}-cos C~~~...(1)\\A+B+C=180\\A+B=180-C\\\frac{A+B}{2}=90-\frac{C}{2}\\cos \frac{A+B}{2}=cos(90-\frac{C}{2})=sin \frac{C}{2}\\cos C=1-2sin^2\frac{C}{2}\\(1)=2 sin \frac{C}{2} cos \frac{A-B}{2}-1+2sin^2\frac{C}2}\\=2sin\frac{C}{2}[cos \frac{A-B}{2}+sin \frac{C}{2}]-1~~~...(2)\\\\now~again~A+B+C=180\\C=180-(A+B)\\sin\frac{C}{2}=sin(90-\frac{A+B}{2})=cos \frac{A+B}{2}\\(2)=2sin\frac {C}{2}[cos \frac{A-B}{2}+cos \frac{A+B}{2}]-1\\$

$=-1+2sin\frac{C}{2}*2cos \frac{\frac{A-B}{2} +\frac{A+B}{2} }{2} cos \frac{\frac{A-B}{2} -\frac{A+B}{2} }{2} \\=-1+4sin\frac{C}{2} cos \frac{A}{2} cos\frac{-B}{2} \\=-1+4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}\\(cos(-B)=cos B)$

8 0

4 years ago

Explain how an estimate helps you place the decimal point when multiplying 3.9 × 5.3

natali 33 [55]

An estimate helps you because you round 3.9 to 4 and 5.3 to 5 so then you can give a rough estimation

4 0

3 years ago

Read 2 more answers