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saveliy_v [14]
2 years ago
7

Max rides his scooter toward Kim and then passes her at a constant speed. His distance in feet, d, from Kim t seconds after he s

tarted riding his scooter is given by d = |150 – 9t|. What does the 150 in the equation represent? What does the 9 in the equation represent?
Mathematics
2 answers:
MrRa [10]2 years ago
5 0
What does the 150 in the equation represent?
Max's starting distance from Kim

What does the 9 in the equation represent?
Max's constant speed
sasho [114]2 years ago
4 0

The equation of d=|150-9t| represents the distance from Max to Kim.

<u><em>If you look at the equation, you can say that each second, 9 feet is being lessened from 150. So that means Kim is initially 150 feet ahead of Max and he is closing in at 9 meters per second.</em></u>

  • 150 in the equation represents the initial distance between Max and Kim. 150 is how far Max is from Kim before he starts.
  • 9 in the equation represents the rate at which Max is closing in on Kim. 9 feet per second is Max's rate.

ANSWER:

150 represents the distance Max is from Kim

9 represents that rate at which Max is riding his scooter (<em>9 feet per second</em>)

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Luis made some trail mix .He mixed 4 2/3 cups of popcorn ,1 1/4 cups of peanuts ,1 1/3 cups of raisins ,and 3/4 cup of sunflower
erastova [34]
I would say the answer is A.
4 0
2 years ago
Which represents the polynomial written in standard form?
pshichka [43]

Answer:

Option A.

Step-by-step explanation:

In order to write any polynomial in standard form, we need to check the degree of each term, then write each term in order of degree, from highest to lowest, left to right.

The given polynomial is

8x^2y^2-\dfrac{3x^3y}{2}+4x^4-7xy^3

Here, the combine degree of x and y in each term is 4.

If we arrange the terms according to the degree of x, then

4x^4-\dfrac{3x^3y}{2}+8x^2y^2-7xy^3

If we arrange the terms according to the degree of y, then

-7xy^3+8x^2y^2-\dfrac{3x^3y}{2}+4x^4

Hence, the correct option is A.

3 0
2 years ago
Read 2 more answers
Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an
Charra [1.4K]

1. Consider square ABCD. You know that  

A_{ABCD}=AD^2=200,

then

AB=BC=CD=AD=10\sqrt{2}.

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.

3. Consider trapezoid BFGC. Its area is

A_{BFGC}=\dfrac{FG+BC}{2}\cdot h, where h is the height of trapezoid and is equal to half of AB. Thus,

A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.

4.

A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.

5. Note that angles EGC and CGD are supplementary and

\sin \angle CGD=\sin \angle EGC.

Then

A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.

Answer: A_{CGD}=41.

8 0
2 years ago
The population can be modeled by P(t) = 82.5 − 67.5cos⎡ ⎣(π/6)t ⎤ ⎦, where t is time in months (t = 0 represents January 1) and
Fed [463]

Answer:

The intervals in which the population is less than 20,000 include

(0 ≤ t < 0.74) and (11.26 < t ≤ 12)

Step-by-step explanation:

P(t) = 82.5 - 67.5 cos [(π/6)t]

where

P = population in thousands.

t = time in months.

During a year, in what intervals is the population less than 20,000?

That is, during (0 ≤ t ≤ 12), when is (P < 20)

82.5 - 67.5 cos [(π/6)t] < 20

- 67.5 cos [(π/6)t] < 20 - 82.5

-67.5 cos [(π/6)t] < -62.5

Dividing both sides by (-67.5) changes the inequality sign

cos [(π/6)t] > (62.5/67.5)

Cos [(π/6)t] > 0.9259

Note: cos 22.2° = 0.9259 = cos (0.1233π) or cos 337.8° = cos (1.8767π) = 0.9259

If cos (0.1233π) = 0.9259

Cos [(π/6)t] > cos (0.1233π)

Since (cos θ) is a decreasing function, as θ increases in the first quadrant

(π/6)t < 0.1233π

(t/6) < 0.1233

t < 6×0.1233

t < 0.74 months

If cos (1.8767π) = 0.9259

Cos [(π/6)t] > cos (1.8767π)

cos θ is an increasing function, as θ increases in the 4th quadrant,

[(π/6)t] > 1.8767π (as long as (π/6)t < 2π, that is t ≤ 12)

(t/6) > 1.8767

t > 6 × 1.8767

t > 11.26

Second interval is 11.26 < t ≤ 12.

Hope this Helps!!!

3 0
2 years ago
Jason ran 5/7 of the distance around the school track. Sara ran 4/5 of Jason's distance. What fraction of the total distance aro
masha68 [24]
Jason: 5/7
Sara: 4/5

Think of it this way: if the total track is a mile, and Jason runs 5/7 of it,  he has run 5/7 of a mile. Then Sara runs 4/5 OF 5/7, with "of" meaning "times," so she runs 4/5 x 5/7, which gives you 20/35, which simplifies to 4/7. You can also think of Sara's distance as 80 percent of Jason's. If she runs "80 percent of the sevenths that Jason ran," that means she ran 4 out of his 5 sevenths.
7 0
2 years ago
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