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PolarNik [594]
3 years ago
13

A runner covers 55 yards in 5.5 seconds. Find the runner's average speed in miles per hour.

Mathematics
1 answer:
marusya05 [52]3 years ago
7 0
V=\frac{s}{t},where\ V-speed,\ s-distence\ ,t-time\\\\V=\frac{55y}{5,5s}=10\frac{y}{s}=10\frac{0,000568m}{\frac{1}{3600}h}=\frac{0,00568m}{\frac{1}{3600}h}=0,00568\cdot3600\frac{m}{h}=\\\\=20,448\frac{m}{h}
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3 years ago
I don't care about getting the answers but rather a step-by-step explanation for this problem: A simple interest, 8-month loan o
alisha [4.7K]

Answer:

Step-by-step explanation:

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

From the information given

T = 8 months = 8/12 = 2/3 years

P = $3000

R = 9.3%

Therefore

I = (3000 × 9.3 × 2/3)/100

I = 18600/100

I = $186

The maturity value (in dollars) of this loan would be

3000 + 186 = $3186

7 0
3 years ago
Find the zeros of the function. Enter the solutions from least to greatest . h(x) = (- 4x - 3)(x - 3) lesser x = greater
icang [17]

9514 1404 393

Answer:

  -3/4, 3

Step-by-step explanation:

The zeros are the values of x that make h(x)=0. Those are the values of x that make the factors of h(x) be zero.

  -4x -3 = 0   ⇒   x = -3/4

  x -3 = 0   ⇒   x = 3

The zeros of the function are -3/4 and 3.

3 0
3 years ago
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
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<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
8 0
3 years ago
Please Help Me:!!!!!
butalik [34]

Answer:

Vertical stretch of factor of 3

translated to the right 2

translated up 5

Step-by-step explanation:

6 0
3 years ago
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