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2 years ago

What is 1,995,298 rounded to the nearest 10,000

2 answers:

6 0

Salutations!

What is 1,995,298 rounded to the nearest 10,000?

To round the number to the nearest 10,000, you need to know whether the number next 10,000 place is greater than 5 or lesser than 5. If its greater, you need to round up. If its lesser than 5, you need to round down.

1,995, 298 ≈ 2,000,000.

Hope I helped (:

Have a great day!

iren [92.7K]2 years ago

4 0

the answer is 2,000,000

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What is the 101st term in the sequence 997, 989, 981, ...?

melamori03 [73]

Answer: first option.

Step-by-step explanation:

Find the common difference d of the arithmetic sequence:

$d=a_n-a_{(n-1)}\\d=989-997\\d=-8$

Then the formula for the 101st term is the shown below:

$a_n=a_1+(n-1)d$

Where:

$a_1=997\\d=-8\\n=101$

Substitute values into the formula. Therefore, you obtain:

$a_n=997+(101-1)(-8)=197$

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2 years ago

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larisa86 [58]

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3 years ago

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3 years ago

A manufacturer wants to make open tin boxes from pieces of tin with dimensions 8 in. by 15 in. by cutting equal squares from the

Stells [14]

Answer:

$\frac{5}{3}$ in

Step-by-step explanation:

Let x be the side of square.

Length of box=8-2x

Width of box=15-2x

Height of box=x

Volume of box= $L\times B\times H$

Substitute the values then we get

Volume of box=V(x)= $(8-2x)(15-2x)x=(15x-2x^2)(8-2x)$

$V(x)=120x-30x^2-16x^2+4x^3$

$V(x)=4x^3-46x^2+120x$

Differentiate w.r.t x

$V'(x)=12x^2-92x+120$

$V'(x)=0$

$12x^2-92x+120=0$

$3x^2-23x+30=0$

$3x^2-18x-5x+30=0$

$3x(x-6)-5(x-6)=0$

$(x-6)(3x-5)=0$

$x-6=0\implies x=6$

$3x-5=0\implies x=\frac{5}{3}$

Again differentiate w.r.t x

$V''(x)=24x-92$

Substitute x=6

$V''(6)=24(6)-92=52>0$

Substitute x=5/3

$V''(5/3)=24(5/3)-92=-52$

Hence, the volume is maximum at x= $\frac{5}{3}$

Therefore, the side of the square , $x=\frac{5}{3}$ in cutout that gives the box the largest possible volume.

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2 years ago

Identify the vertex, the axis of symmetry, the maximum or minimum value, and the range of the parabola for y=x^2+6x+13

charle [14.2K]

Answer:

Step-by-step explanation:

First we can determine the x value of our vertex via the equation:

$x=\frac{-b}{2a}$

Note that in general a quadratic equation is such that:

$ax^2+bx+c$

In this case a,b and c are the coefficients and so a=1, b=6 and c=13.

Therefore we can determine the x component of the vertex by plugging in the values known and so:

$x=\frac{-b}{2a}=\frac{-6}{2(1)}=\frac{-6}{2}=-3$

Now we can determine the y-component of our vertex by plugging in the x-component to the equation and so:

$f(x)=x^2+6x+13\\\\f(-3)=(-3)^2+6(-3)+13\\\\f(-3)=4$

Therefore our vertex is (-3,4). Now in vertex our x component determines is the axis of symmetry so the equation for axis of symmetry is:

x=-3

Similarly, the y-component of our vertex is the minimum or maximum. In this case it is the minimum you can determine this because a is positive meaning that the parabola will point up, and so the equation for the minimum is:

y=4

The range of the formula is the smallest y-value meaning the minimum y=4 and all real numbers that are more than 4, mathematically:

Range = All real numbers greater than or equal to 4.

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3 years ago