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2 years ago

Can someone please solve this

Mathematics

2 answers:

Svetradugi [14.3K]2 years ago

4 0

Answer:

The answer is the 3rd one.

Step-by-step explanation:

3 times 3 is 9

4 times 4 is 16

13 is in between those 2 numbers

Korolek [52]2 years ago

4 0

Answer:

Step-by-step explanation:

square root of 13 is 3.60555127546, so it would be the third one

can u think abut brainlist

3^2 : 9

4^2 : 16

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Find the missing length in the triangle. Round to the nearest tenth.

Alik [6]

That depends on what kind of triangle it is. If it is a right triangle, then which sides are given values?

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2 years ago

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The dwarf seahorse hippocampus zosterae swims at a rate of 52.68 feet per hour.convert this speed to ines per minute

DaniilM [7]

<h3>52.68 feet per hour = 10.536 inches per minutes</h3>

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem !

$1 ~ foot = 30.48 ~ cm$

$1 ~ inch = 2.54 ~ cm$

$1 ~ foot = 12 ~ inches$

$\boxed {52.68 ~ feet = 52.68 \times 12 ~ inches = 632.16 ~ inches}$

$1 ~ day = 24 ~ hours$

$\boxed {1 ~ hour = 60 ~ minutes}$

If the dwarf seahorse hippocampus zosterae swims at a rate of 52.68 feet per hour , then :

$52.68 ~ feet/h = \frac{52.68 ~ feet}{1 ~ hour}$

$52.68 ~ feet/h = \frac{52.68 \times 12 ~ inches}{60 ~ minutes}$

$52.68 ~ feet/h = \frac{632.16 ~ inches}{60 ~ minutes}$

$\large {\boxed {52.68 ~ feet/h = 10.536 ~ inches/min} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

3 0

3 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.