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3 years ago

Solve the linear inequality 10x+18<−2

Mathematics

2 answers:

asambeis [7]3 years ago

7 0

The answer I got:

x<-2

igor_vitrenko [27]3 years ago

3 0

10x+18∠-2
-18 -18

10x∠-20
---- -----
10 10

x∠-2

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The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

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3 years ago

Question 12<br> With steps please

34kurt

$Answer:\ \boxed{5^2}$

$\sqrt[n]{a}=a^\frac{1}{n}\\\\\text{therefore}\\\\\sqrt5=5^\frac{1}{2}\\----------------\\25=5^2\\----------------\\78,125=5^7\\----------------\\\text{use}\\\\a^n\cdot a^m=a^{n+m}\\\\\dfrac{a^n}{a^m}=a^{n-m}\\----------------\\\\\dfrac{\sqrt5\times78,125}{25\times5^{\frac{7}{2}}}=\dfrac{5^\frac{1}{2}\times5^7}{5^2\times5^\frac{7}{2}}=\dfrac{5^{\frac{1}{2}+7}}{5^{2+3\frac{1}{2}}}=\dfrac{5^{7\frac{1}{2}}}{5^{5\frac{1}{2}}}=5^{7\frac{1}{2}-5\frac{1}{2}}=\boxed{5^2}$

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3 years ago

A new sidewalk will be 6 feet wide, 100 feet long, and filled in a debt of 9 inches with concrete. How many cubic yards of concr

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Answer: 16.67 yd^3 cubic yards

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3 years ago

Read 2 more answers

I need both of these help!

marysya [2.9K]

The angle next to the angle 5x is 180 - 5x

All angles sum to 360

(6x - 58) + (2x + 4) + (180 - 5x) = 180
3x + 126 = 180
3x = 54
x = 18

For question 10, sum of angles in pentagon is 540 degrees

x + 105 + 85 + 114 + 126 = 540
x + 430 = 540
x = 110

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3 years ago

Craig solicited donations for his school's jump-rope a-thon. He collected $13.00 in fixed donations and pledges totaling $1.50 f

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Answer:

Step-by-step explanation:

13+1.50m

13+1.50(8)

13+12

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2 years ago