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3 years ago

10

Evaluate mathematical problem for one by two

itle=" 1000 \times 10" alt=" 1000 \times 10" align="absmiddle" class="latex-formula">

2 answers:

svp [43]3 years ago

7 0

1000*10

Multiply values together

10000

serious [3.7K]3 years ago

7 0

1000*10=10000. Just add 1 zero to the back because 10 has 1 zero.

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5. A soccer ball is kicked from the ground level with an upward velocity of 60 feet per second. The

elixir [45]

Answer:

-16(1)^2+60(1)= -16+60=44

Step-by-step explanation:

6 0

4 years ago

First, Khianna must walk from the front gate to the Ferris wheel. This distance is 2.5 cm on Khianna's map. How far is the actua

spin [16.1K]

Okie doke. 1cm on the map equals 500 meters. The distance from the gate to the ferris wheel on the map is 2.5cm. To find the actual distance, multiply the length on the map by the distance per centimeter. 500 * 2.5 is 1,250. The actual distance is 1,250 meters.

Note: When it comes to problems like these, you must always multiply the distance on the map by the scale unit (in other words, the actual distance on how far it is per the smaller unit).

7 0

3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

4 0

3 years ago

Read 2 more answers

Please help immediately

Burka [1]

Answer: $f'(1)=\dfrac{4}{2y+1}$

<u>Step-by-step explanation:</u>

To find the tangent, you need to find the derivative with respect to x.

Then substitute x = 1 into the derivative.

Given: 0 = 2x² - xy - y²

Derivative: 0 = 4x - y' - 2yy'

Solve for y': 2yy' + y' = 4x

y'(2y + 1) = 4x

$y'=\dfrac{4x}{2y+1}$

$\text{Substitute x = 1:}\quad y'(1)=\dfrac{4}{2y+1}$

8 0

3 years ago

How does the volume of an oblique cylinder change if the radius is reduced to 2/9 of its original size and the height is quadrup

vesna_86 [32]

Volume of a cylinder = π r² h

Let us assume the following values:
radius = 9
height = 10

Volume = 3.14 * 9² * 10
= 3.14 * 81 *10
Volume = 2,543.40

Changes:
radius is reduced to 2/9 of its original size = 9 x 2/9 = 2
height is quadrupled = 10 x 4 = 40

Volume = π r² h
= 3.14 * 2² * 40
= 3.14 * 4 * 40
Volume = 502.40

Original volume = 2543.40 V.S. Volume after change = 502.40

The volume of an oblique cylinder decreased when its radius was decreased to 2/9 of its original size and its height is increased 4 times.

4 0

3 years ago