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MA_775_DIABLO [31]
3 years ago
13

What is the value of x2+9x+20 when x could be -5, -3 or -6

Mathematics
1 answer:
ANTONII [103]3 years ago
3 0

Answer:

Step-by-step explanation:

You should use a ^ to indicate an exponent. x^2 is the same as x².

x² + 9x + 20

Replace the x with (-5).

(-5)² + 9(-5) + 20 = 25-45+20 = 0

Then do the same with x = (-3):

(-3)² + 9(-3) + 20 = 9 - 27 + 20 = 2

You can do x = (-6)

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I need help ASAP 30 points+brainleist
yulyashka [42]

Answer:

A) 40 feet.

B) 7 inches.

Step-by-step explanation:

A) To acquire 8 inches, multiply the corresponding scale factor by 8:

1 inch equals 5 feet; 8 inches equals 40 feet

The genuine house stands 40 feet tall.

B) You can write the percentage if you want...

dwg / (35 feet) = 1 in / (5 ft)

When you multiply by 35 feet, you get...

7 in dwg = (1 in)(35 ft)/(5 ft)

In the scale drawing, the home measures <em>7 inches in length</em>.

5 0
1 year ago
Read 2 more answers
What is the solution of the inequality shown below X-3&gt;7
True [87]

Step-by-step explanation:

x-3>7 =>

x>10

hope this helps

5 0
3 years ago
Alyssa spent 9.20 for 16 packs of gum at this rate how much would it cost alyssa to buy 30 packs of gum.
zhuklara [117]

$17.25. Divide 9.20 by 16 and mulitply by 30.

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3 years ago
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What is the solution to -3x&lt;9
GaryK [48]
-3x<9. The way to figure this out is to divide by -3 to get x by itself. Remember, when dividing by a negative, you need to flip the inequality sign.
x>-3 is your answer.
I hope this helps!
7 0
3 years ago
A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are
skelet666 [1.2K]

A) The maximum revenue is 450000$

B) The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

C) When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

A)p(x)=300−(x/20​),

revenue R(x)=p*x

revenue R(x)=300x -(x2/20)

for maximum revenue dR/dx =0 ,

=>300-(2x/20)=0

=>x/10=300

=>x=3000

maximum revenue = R(3000)=300*3000 -(30002/20)

maximum revenue = R(3000)=450000$

B) profit =revenue -cost

profit P(x)=300x -(x2/20)-72000-60x

profit P(x)=240x -(x2/20)-72000

for maximum cost dP/dx =0

240 -(2x/20)=0

x=240*10

x=2400

p(2400)=300−(2400/20​)=180

profit P(2400)=240*2400 -(24002/20)-72000 =216000

The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

c)

profit =revenue -cost -tax

profit P(x)=300x -(x2/20)-72000-60x-55x

profit P(x)=185x -(x2/20)-72000

for maximum cost dP/dx =0

185-(2x/20)=0

x=185*10

x=1850

p(1850)=300−(1850/20​)=207.5

profit P(1850)=185*1850 -(18502/20)-72000

profit P(1850)=99125$

When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

To know more about maximum profit check the below link:

brainly.com/question/4166660

#SPJ4

5 0
10 months ago
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