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34kurt
3 years ago
12

Solve the equation. x/5 +9 = 4

Mathematics
1 answer:
solniwko [45]3 years ago
6 0

Mult. all 3 terms by 5: x + 45 = 20.

Subtr. 45 from boths ides: x = -25 (answer)

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? x (3 + 2) = 12 + 8 What's the number that goes into the question mark?​
andrew11 [14]

Answer:

?=4

Step-by-step explanation:

?*(3+2)=12+8

?(5)=20

20/5=4

?=4

4*(5)=20

20=20

7 0
3 years ago
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Find the slope given: (11, 12) &amp; (1,8)<br> m =
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12/20 hope this help you
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13. Solve the following:<br> a. 2.68 - 1.44 =<br> b. 5.05 X 0.04 =<br> c. 5.144 = 1.6 =
AURORKA [14]
A) 1.24 B) 0.202 and C) 3.215.
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3 years ago
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A pilot is flying a CRJ-900 from Phoenix, Arizona, to Louisville, Kentucky. The aircraft cannot take off or land if the temperat
ioda

We need to find an inequality that represents the possible landing or take of conditions of an aircraft, the required graph and one case where the inequality will be tested.

The required inequality is -40.

The graph represents the allowable temperature range in which the aircraft can takeoff or land.

The pilot could not have taken off on June 1990, the temperature in Phoenix, Arizona.

Let the temperature be denoted by x

The temperature cannot be at or below -40^{\circ}\text{F}, so

x>-40

The temperature cannot be at or above 118^{\circ}\text{F}, so

x

The inequality will be -40.

In the graph it can be seen that all points between -40 and 118 falls in the shaded region but it will exclude the points -40 and 118.

The graph represents the allowable temperature range in which the aircraft can takeoff or land.

The temperature in June 1990 in Phoenix, Arizona was 122^{\circ} which is more than 118^{\circ}\text{F}.

Hence, the pilot could not have taken off on this day.

Learn more:

brainly.com/question/20383699

brainly.com/question/11612965

4 0
2 years ago
Find the derivative of the function f(x) = (x3 - 2x + 1)(x – 3) using the product rule.
julsineya [31]

Answer:

Step-by-step explanation:

Hello, first, let's use the product rule.

Derivative of uv is u'v + u v', so it gives:

f(x)=(x^3-2x+1)(x-3)=u(x) \cdot v(x)\\\\f'(x)=u'(x)v(x)+u(x)v'(x)\\\\ \text{ **** } u(x)=x^3-2x+1 \ \ \ so \ \ \ u'(x)=3x^2-2\\\\\text{ **** } v(x)=x-3 \ \ \ so \ \ \ v'(x)=1\\\\f'(x)=(3x^2-2)(x-3)+(x^3-2x+1)(1)\\\\f'(x)=3x^3-9x^2-2x+6 + x^3-2x+1\\\\\boxed{f'(x)=4x^3-9x^2-4x+7}

Now, we distribute the expression of f(x) and find the derivative afterwards.

f(x)=(x^3-2x+1)(x-3)\\\\=x^4-2x^2+x-3x^3+6x-4\\\\=x^4-3x^3-2x^2+7x-4 \ \ \ so\\ \\\boxed{f'(x)=4x^3-9x^2-4x+7}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

6 0
3 years ago
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