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Veseljchak [2.6K]
3 years ago
6

Solve.

Mathematics
1 answer:
yarga [219]3 years ago
5 0
- 3/5x + 15 > 720
- 3/5x > 705
X < -423
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15 divided by 3.15 what is the unit rate
stiv31 [10]

Answer: 4.8 i think

Step-by-step explanation: its just division

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3 years ago
What is the third quartile of this data set? *I NEED THE HELP ASAP* thx
diamong [38]

I think the answer is 44

7 0
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Two angles in triangle PQR are congruent, ∠P and ∠Q; ∠R measures 28.45°. What is the measure of ∠P?
Ymorist [56]

Step-by-step explanation:

the sum of all angles in a triangle is ALWAYS 180°.

and by using the word "congruent" : for angles that only means they are equally large.

we know the third angle.

so,

180 = P + Q + R = 2×P + 28.45

(because P = Q)

2×P = 151.55

P = Q = 151.55 / 2 = 75.775°

5 0
1 year ago
Read 2 more answers
Find the parabola through
Anni [7]

Answer:

y = - 3x² - 24x - 60

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (- 4, - 12 ), thus

y = a(x + 4)² - 12

To calculate a substitute (- 7, - 39) into the equation

- 39 = a(- 7 + 4)² - 12 ( add 12 to both sides )

- 27 = 9a ( divide both sides by 9 )

- 3 = a

y = - 3(x + 4)² - 12 ← in vertex form

Expand (x + 4)²

y = - 3(x² + 8x + 16) - 12

   = - 3x² - 24x - 48 - 12

y = - 3x² - 24x - 60 ← in standard form

   = - 3(x²

3 0
3 years ago
Let f(x) = xe^-x+ ce^-x, where c is a positive constant. For what positive value of c does f have an absolute
Illusion [34]

Answer:

c=6

Step-by-step explanation:

The absolute maximum of a continuous function f(x) is where f'(x)=0. Therefore, we must differentiate the function and then set x=-5 and f'(x)=0 to determine the value of c:

f(x)=xe^{-x}+ce^{-x}

f'(x)=-xe^{-x}+e^{-x}-ce^{-x}

0=-(-5)e^{-(-5)}+e^{-(-5)}-ce^{-(-5)}

0=5e^{5}+e^{5}-ce^{5}

0=e^5(5+1-c)

0=6-c

c=6

Therefore, when c=6, the absolute maximum of the function is x=-5.

I've attached a graph to help you visually see this.

7 0
3 years ago
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