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Svetlanka [38]
2 years ago
9

Can someone plz help me

Mathematics
2 answers:
lidiya [134]2 years ago
4 0

Answer:

The answer is C. <u>5</u><u><em>a² - </em></u><u>4</u><u><em>a</em></u><u> + 1 </u>

<h3>Step- by- Step:</h3>

<em>First you want to Distribute the Negative Sign:</em>

= 10a² + 3 − 5a + −1 (2 + 5a² − a)

= 10a² + 3 + −5a + (−1) (2) + −1 (5a²) + −1 (−a)

= 10a² + 3 + −5a + −2 + −5a² + a

<em>Add Like Terms:</em>

= 10a² + 3 + −5a + −2 + −5a² + a

= (10a² + −5a²) + (−5a + a )+( 3+ −2)

<u>= 5a² + −4a + 1</u>

<h3>Happy To Help! :)</h3>

Allisa [31]2 years ago
3 0

Answer:

i think its c

Step-by-step explanation:

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select the values that make the inequality h/4 greater than or equal to -1 true then write an equivalent inequality in terms of
jok3333 [9.3K]

Answer:

Values that make it equal:

-4, -3, -1, 0, 4

Written equivalent inequality:

h≥−4

Step by Step for the written inequality:  

h/4 ≥−1

Multiply both sides by 4. Since 4 is positive, the inequality direction remains the same.

h≥−4

I hope this helps you!! :)

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2 years ago
Point A is located at (-5, 2) on a coordinate grid. Point A is translated 4 units to the right and 2 units down to create point
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The correct solution is A thanks
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If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :    

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

(\text{Average velocity})_{0.1\ s}=696\ ft/s

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

5 0
3 years ago
A circular plate has a diameter of 11 inches. Which is closest tothe area of this plate? Show your work.A 17.3 square inchesC 95
monitta
Area of a circle= πr²

Data:
diameter=11 in

ratius=diameter/2=11 in/2=5.5 in

Area of this circular plate=π(5.5 in)²=30.25π in²=30.25 * 3.141592... in²≈
≈95.0 in².

answer :  c  95.0 in²
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The 0 is in the ten thousands
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