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Leno4ka [110]
3 years ago
12

Brandy jogged 2 1/2 Mile in 30 minutes 3 1/2 miles in 3/4 hour who jogged at a slower rate

Mathematics
1 answer:
Artyom0805 [142]3 years ago
6 0

Answer: The second person


Step-by-step explanation:


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The circumference of a circle is 31.4 centimeters. what is the area of the circle?
icang [17]

Answer:  197.292

Step-by-step explanation:

4 0
3 years ago
Which is the best estimate of 174% of 141?
xeze [42]
245.34 is 174% of 141 :)
(1.74*141=245.34)

It helps to remember that "of" in math means to multiply. That's helped me for years.
4 0
3 years ago
Read 2 more answers
(1 point) If p(x) and (x) are arbitrary polynomials of degree at most 2, then the mapping =p(-1)q(-1) + p(070) +p(3)q(3) defines
Ainat [17]

If p(x) and q(x) are arbitrary polynomials of degree at most 2 then

||p||||q|| = 26(\sqrt{640}) and angle between p(x) and q(x) is 0.233.

Given that

<p,q> = p(-1)q(-1) + p(0) q(0) + p(3)q(3)

and p(x) = 2x²+ 6 , q(x)= 4x²-4x

then the values of p and q at x = -1,0,3 are given as;

x = -1,

p(-1) = 2(-1)² + 6 = 8   ,    q(-1) = 4(-1)² - 4(-1) = 8

x = 0,

p(0) = 2(0)² + 6 = 6   ,    q(0) = 4(0)² - 4(0) = 0

x = 3,

p(3) = 2(3)² + 6 = 24  ,  q(3) = 4(3)²- 4(3) = 24.

<p,q> = p(-1)q(-1) + p(0)q(0) + p(3)q(3)

         = (8)(8) + (6)(0) + 24(24)

         = 64 + 0 + 576

<p,q> = 640

Now we have to find ||p|| ||q||, for this we'll find ||p|| and ||q||

||p|| = \sqrt{ < p,p > }

     = \sqrt{8(8) + 6(6) + 24(24)}

     = \sqrt{676}

||p|| = 26

and

||q|| = \sqrt{ < q,q > }

      =\sqrt{8(8) + 0(0) + 24(24)}

||q||  =\sqrt{640}

∴||p||||q|| = 26(\sqrt{640\\)

Now we have to find angle between p(x) and q(x),

∴ α = cos⁻¹\frac{ < p,q > }{||p||||q||}

      = cos ⁻¹ \frac{640}{26(\sqrt{640}) }

      = cos ⁻¹ \frac{4\sqrt{10} }{13}

  α  = 13.34°

In radian

α = 0.233.

To know more about Inner product here

brainly.com/question/14185022

#SPJ4

5 0
1 year ago
9. A ladder of length 23 feet leans against the side of a building. The angle of elevation of the ladder is 76. Find the distanc
irakobra [83]

<h3><u>Answer:</u></h3>

  • B) 22.32 feet

\quad\rule{300pt}{1pt}

<h3><u>Solution</u><u>:</u></h3>

we are given that , a ladder is placed against a side of building , which forms a right angled triangle . We wre given one side of a right angled triangle ( hypotenuse ) as 23 feet and the angle of elevation as 76 ° . We can find the Perpendicular distance from the top of the ladder go to the ground by using the trigonometric identity:

\qquad\quad\bull ~{\boxed{\bf{ Sin\theta =\dfrac{Perpendicular}{Hypotenuse}}} }~\bull

Here,

  • hypotenuse = 23 feet
  • \theta = 76°
  • Value of Sin\theta = 0.97
  • Perpendicular = ?

\quad\dashrightarrow\quad \sf { sin\theta =\dfrac{P}{H}}

\quad\dashrightarrow\quad \sf { sin76° =\dfrac{P}{23}}

\quad\dashrightarrow\quad \sf { 0.97=\dfrac{P}{23}}

\quad\dashrightarrow\quad \sf {P = 0.97 \times 23 }

\quad\dashrightarrow\quad \sf { P = 22.32}

‎ㅤ‎ㅤ‎ㅤ~<u>H</u><u>e</u><u>n</u><u>c</u><u>e</u><u>,</u><u> </u><u>the </u><u>distance </u><u>from </u><u>the </u><u>top </u><u>of </u><u>the </u><u>ladder </u><u>to </u><u>the </u><u>ground </u><u>is </u><u>2</u><u>2</u><u>.</u><u>3</u><u>2</u><u> </u><u>feet </u><u>!</u>

\rule{300pt}{2pt}

4 0
3 years ago
During a promotional weekend, a state fair gives a free admission to every 179th person that enters the fair. On Saturday, there
V125BC [204]

Answer:

89

Step-by-step explanation:

Given that,

During a promotional weekend, a state fair gives a free admission to every 179th person that enters the fair.

No of people attending the fair on Saturday is 8,633 amd No of people attending the fair on Sunday is 7,400.

We need to find the no of people that received a free admission over the two days.

Dividing 8,633 by 179 gives 48 as quotient and 41 as remainder. It means on Saturday 48 people entered for free.

Dividing 7,400 by 179 gives 41 as quotient and 61 as remainder. It means on Sunday 41 people entered for free.

Total no of people,

T = 48 + 41

T = 89

Hence, there are 89 people for free entries.

6 0
4 years ago
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