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2 years ago

3) Write the inequality that

Mathematics

1 answer:

masha68 [24]2 years ago

3 0

Answer:

See below

Step-by-step explanation:

Owing represent a negative balance.

So the inequality for this case is:

- 26.00 > - 147.00

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Help me with these math questions....

Nady [450]

Answer: cotθ

Step-by-step explanation:

tanθ * cos²θ * csc²θ

= $\dfrac{sin\theta}{cos\theta} * \dfrac{cos\theta*cos\theta}{} *\dfrac{1}{sin\theta*sin\theta}$

= $\dfrac{cos\theta}{sin\theta}$

= cotθ

Answer: B

Step-by-step explanation:

The parent graph is y = x²

The new graph y = -x² + 3 should have the following:

reflection over the x-axis
vertical shift up 3 units

Answers:

a. Quadrant II
b. negative
c. $\dfrac{\pi}{6}$
d. C
e. $-\dfrac{\sqrt{3}}{3}$

Explanation:

$\dfrac{17\pi}{6} - \dfrac{12\pi}{6} = \dfrac{5\pi}{6}$

a) Quadrant 2 is: $\dfrac{\pi}{2} < \theta < \pi$

b) In Quadrant 2, cos is negative and sin is positive, so tan is negative

c) $\pi-\dfrac{5\pi}{6}$ = $\dfrac{\pi}{6}$

d) the reference line is above the x-axis so it is negative --> $-tan\dfrac{\pi}{6}$

e) $tan(\dfrac{5\pi}{6})=\dfrac{1}{-\sqrt{3}}=-\dfrac{\sqrt{3}}{3}$

4 0

2 years ago

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

2 years ago

Is the square number of 32 a whole number?

Aneli [31]

No it’s not the square root of 32 is 5.65685424949

4 0

3 years ago

Read 2 more answers

I feel like I am asking so my questions but I dont really care its a Sunday xD

dimulka [17.4K]

You would add like terms so it would be 1x then you add 1 to 7 so you would get 8.. the you divide both sides by 1x and you'd get 8=x

4 0

2 years ago

Read 2 more answers

How do you do this i totaly forgot.

vagabundo [1.1K]

Step-by-step explanation:

7^2+8^2=113

Then square root 113 which is 10.63 to 2.d.p

So x=10.63

5 0

2 years ago