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4 years ago

Solve the inequality for x. –3x – 3 < –63

Mathematics

2 answers:

arlik [135]4 years ago

7 0

Answer:

ill help:)

Step-by-step explanation:

Morgarella [4.7K]4 years ago

7 0

Answer:

x < 20.

Step-by-step explanation:

-3x-3<-63

+3 +3

-3x<-60

÷-3. ÷-3

x<20

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A firework is fired into the air from the top of a barn. Tis hight (h) above the ground in yards after (t) seconds is given by t

Grace [21]

The given function for the height of the firework is a quadratic function

1. Time at which the firework reaches the maximum height is 1 seconds

2. The maximum height of the firework, is 25 yards

3. Time after which the firework will fall to the ground is (1 + √5) seconds

Reason:

The given function that represents the height of the fireworks with time is

presented as follows;

h(t) = -5·t² + 10·t + 20

1. The time at which the firework reaches its maximum height is given by

the maximum point of the given function as follows;

The x-value of the maximum point of a quadratic function is $x = \dfrac{b}{2 \cdot a}$

Where;

a, and b, are the coefficient of x² and x, in the general form of a quadratic function f(x) = a·x² + b·x + c

By comparison, we have;

$t = -\dfrac{10}{2 \times (-5)} = 1$

The time at which the firework reaches the maximum height is t = 1 seconds

2. The maximum height is given by plugging in the value of t, at the maximum point into the given function as follows;

h(1) = -5×1² + 10×1 + 20 = 25

The maximum height of the firework, f(1) = 25 yards

3 The time at which the firework will fall to the ground, is given by the zero of the function as follows;

When the firework falls to the ground, h(t) = 0 = -5·t² + 10·t + 20

Dividing both sides by (-5) gives;

$\dfrac{0}{-5} = \dfrac{ -5 \cdot t^2 + 10 \cdot t + 20}{-5} = t^2 - 2 \cdot t - 4$

t² - 2·t - 4 = 0

By the quadratic formula $x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$ , we get;

$t = \dfrac{2\pm \sqrt{(-2)^{2}-4\times 1\times (-4)}}{2\times 1} = \dfrac{2\pm \sqrt{20}}{2\times 1} = \dfrac{2\pm 2 \times \sqrt{5}}{2\times 1} = 1 \pm \sqrt{5}$

Therefore;

The time after which the firework will fall to the ground, t = 1 + √5 seconds

Learn more here:

brainly.com/question/20628403

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3 years ago

peter is buying 4 calculators to donate to his class. each calculator costs $9. he is also buying a $1 bottle of water for himse

Ksivusya [100]

It is at least 40$ that he is going to spend.

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3 years ago

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Common denominators for 5/6 and 8/21 help me!!!!!!!!!

Helga [31]

The common denominator would have to be 42

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3 years ago

When a store had Sold 2/5 of the shirts that were on display, they brought out another 30 from the stockroom. The store likes to

Inga [223]

Answer:

Part A) 3/5 is the total number of shirts remaining on display after selling 2/5

Part B) $x \geq 200$ (All real whole numbers greater than or equal to 200 shirts)

Part C) Initially on display were the amount of 200 or more shirts

Step-by-step explanation:

Let

x-------> total number of shirts on display

we know that

The inequality that represent the situation is

$(1-\frac{2}{5})x+30 \geq 150$

$(\frac{3}{5})x+30 \geq 150$

Part A) Explain what 3/5 means in the inequality

3/5 is the total number of shirts remaining on display after selling 2/5

Part B) Solve the inequality

we have

$(\frac{3}{5})x+30 \geq 150$

Subtract 30 both sides

$(\frac{3}{5})x \geq 150-30$

$(\frac{3}{5})x \geq 120$

Multiply by 5/3 both sides

$x \geq 120(5/3)$

$x \geq 200\ shirts\ on\ display$

Part C) Explain what the solution means in the situation

1) Initially on display were the amount of 200 or more shirts

2) 2/5 were sold and 120 or more shirts were left on display.

3) The store brought out another 30 shirts from the stockroom and placed them on display, for a total of 150 shirts or more on display

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3 years ago

‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️Please please help! I’ve been waiting for 30 minutes and still no one has helped!

Alex777 [14]

PART ;

......Circle A.................
Circumference - 2 pi r
25.12 = 2pi(4)
25.12 = 8i
Divide both sides by 8!!
3.14=pi

...Circle B.....................
Circumference = 2 pi r
9.42 = 2pi(3/2)
9.42 = 3pi
Divide both sides by 3!!!
3.14 = pi

PART B

.............Circle A...............
A=pi r ^2
50.24 = pi(4)^2
50.24=16pi
Divide both sides by 16!!!
3.14=pi

.............Circle B.............
A=pi r^2
7.065=pi(3/2)^2
7.065= 9pi/4
Divide both sides by 9/4!!!
3.14=pi

PART C

They used exactly 3.14 for the value of pi in CIRCLE A and B to get the circumference and the area :D

I hope this makes sense!!!
When I wrote 'pi', it was supposed to be the symbol for pi.

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3 years ago