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4 years ago

An object is launched at 9.8 meters per second from a 73.5 meter tall platform. The object’s

Mathematics

1 answer:

leva [86]4 years ago

7 0

Answer:

Step-by-step explanation:

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I need the volume for 7 and 8

Sergio [31]

For number seven i got 1974m

and for number eight i got 374.13

4 0

3 years ago

Read 2 more answers

Tell whether the situation should be represented by a continuous graph or a discrete graph.

andre [41]

1) Discrete graph

2) Continuous graph

I hope this helps! I would check with someone else first though.

6 0

4 years ago

A box witxh a height (x+5) cm has a square base with side x com. A second box with height (x+2) com has a square base with side

Eva8 [605]

Answer:

x=5.37 cm

Step-by-step explanation:

we know that

The volume of the box is

$V=Bh$

where

B is the area of the base of the box

h is the height of the box

we have that

The area of the base is

$B=x^2\ cm^2$

$h=(x+5)\ cm$

The volume of the Box 1

is equal to

$V_1=x^2(x+5)\ cm^3$

$V_1=(x^3+5x^2)\ cm^3$

we have that

The area of the base is

$B=(x+1)^2\ cm^2$

$h=(x+2)\ cm$

The volume of the Box 2

is equal to

$V_2=[(x+1)^2(x+2)]\ cm^3$

$V_2=[(x^2+2x+1)(x+2)]\ cm^3$

$V_2=(x^3+2x^2+x+2x^2+4x+2)\ cm^3$

$V_2=(x^3+4x^2+5x+2)\ cm^3$

Equate the equation of Volume 1 to the equation of Volume 2

$(x^3+5x^2)=(x^3+4x^2+5x+2)$

$(5x^2)=(4x^2+5x+2)$

$5x^2-4x^2-5x-2=0$

$x^2-5x-2=0$

Solve the quadratic equation by graphing

using a graphing tool

The solution is x=5.37 cm

see the attached figure

8 0

4 years ago

The mathematics department has just received a $200000 donation. The department wants to invest in an annuity so that they can g

docker41 [41]

i think the answer is 6.7

8 0

3 years ago

Find the indicated limit, if it exists.(7 points)

Alecsey [184]

Answer:

Step-by-step explanation:

The left hand limit is when we approach zero from left. We use the function on this domain in finding the limit.

$\lim_{x \to 0^-} f(x)=7-x^2$

$\lim_{x \to 0^-} f(x)=7-(0)^2=7$

The right hand limit is

$\lim_{x \to 0^+} f(x)=10x+7$

$\lim_{x \to 0^+} f(x)=10(0)+7=7$

Since the left hand limit equals the right hand limit;

$\lim_{x \to 0} f(x)=7$

7 0

3 years ago