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slega [8]
3 years ago
6

Amy and her father are playing with a squeaky toy. Amy's father is squeezing the toy in front of Amy. Amy is very excited and re

aches for the toy. Amy's father, however, quickly hides the toy behind his back. At this point, Amy turns away from her father and begins to look at the ladybug design on her dress. Amy is probably approximately what age?
Mathematics
2 answers:
iVinArrow [24]3 years ago
6 0

Answer: 1 year old

Step-by-step explanation:

JulsSmile [24]3 years ago
5 0

Answer:

Amy is 6 moths to 1 year

Step-by-step explanation:

Amy is probably around 6 moths to a year old and the fact that she immediately turns attention to something else tells a lot that she is still too young and hasn't attained much development as a child who has reached 2 years or more.

The reason for her focusing attention somewhere else is due to the fact that her brain hasn't developed much and she doesn't even recall what happened a couple of seconds ago which is one of the characteristics of kids under her age,but with time and she will develop and remember more.

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8×10^7×1.5×3^-1×10^-9​
quester [9]

Answer:

0.04

Step-by-step explanation:

3 0
3 years ago
The area of the sector formed by the 110 degree central angle is 50 units squared. What is the circumference of the circle
Mama L [17]

Answer:

The circumference of the circle is C=45.346 \:units.

Step-by-step explanation:

A sector is the part of a circle enclosed by two radii of a circle and their intercepted arc. A pie-shaped part of a circle.

The area of a circle is given by A_{circle}=\pi r^2

The formula used to calculate the area of a sector of a circle is:

A_{sector}=\frac{central \:angle}{360} \cdot {Area \:of \:whole \:circle}\\\\A_{sector}=\frac{\theta}{360} \cdot \pi r^2

The circumference of a circle is the distance around the outside of the circle and its given by

C=2\pi r

We know the central angle \theta = 110º and the area of the sector 50 units squared.

First, we use the formula to calculate the area of a sector to find the radius.

50=\frac{110}{360} \cdot \pi r^2\\\\\frac{110}{360}\pi r^2=50\\\\\frac{11\pi }{36}r^2=50\\\\11\pi r^2=1800\\\\r^2=\frac{1800}{11\pi }\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\r=\sqrt{\frac{1800}{11\pi }},\:r=-\sqrt{\frac{1800}{11\pi }}

The radius can't be negative. Therefore,

r=\sqrt{\frac{1800}{11\pi }}\approx 7.217 \:units

Next, we apply the formula for the circumference of a circle.

C=2\pi (7.217)=45.346 \:units

7 0
3 years ago
Question from my math test.
Evgen [1.6K]

Answer:

$3.00

Step-by-step explanation:

6 0
3 years ago
Does this converge or diverge? Which of the follow test will be used to figure out the equation?
PSYCHO15rus [73]

From the listed choices, you can prove the series diverges using either the integral or <em>p</em>-series test.

  • Integral test:

We have

\displaystyle\sum_{n=1}^\infty\ge\int_1^\infty\frac{\mathrm dx}{\sqrt x}=\lim_{x\to\infty}2\sqrt x-2

which diverges to infinity, so the series is divergent.

  • <em>p</em>-series test:

The series

\displaystyle\sum_{n=1}^\infty\frac1{n^p}

converges only for p>1. In the given sum, we have p=\frac12, so the series is divergent.

7 0
3 years ago
Positive angles located in the fourth quadrant may be described as___.
gladu [14]

Answer:

Positive angles located in the fourth quadrant may be described as<u> 270≤Ф≤360 .</u>

The option is

4. 270≤Ф≤360

Step-by-step explanation:

When the terminal arm of an angle starts from the x-axis in the anticlockwise direction then the angles are always positive angles.

For Example.

Quadrant I    - 0 to 90°

Quadrant II   - 90° to 180°

Quadrant III  - 180° to 270°

Quadrant IV - 270° to 360° ( 4. 270≤Ф≤360  )

Hence,Positive angles located in the fourth quadrant may be described as<u> 270≤Ф≤360 .</u>

When the terminal arm of an angle starts from the x-axis in the clockwise direction than the angles are negative angles.

Quadrant IV  -          0° to -90°

Quadrant III   -      - 90° to -180°

Quadrant II    -      -180° to -270°

Quadrant I     -      -270° to -360°

5 0
3 years ago
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