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2 years ago

I am thinking of a number that is greater than 142 rounded to the nearest 100 but has a 5 in the ones place! What’s my number!?

Mathematics

1 answer:

Darina [25.2K]2 years ago

6 0

I think your number would be 542

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The quotient of -4 and y

nadya68 [22]

Answer:

T depends on what y equals. -1/4y, i guess

Step-by-step explanation:

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2 years ago

a print shop printed and assembled 8 books in 5 minutes. At this rate how many books can be printed and assembled 3 hours?

ozzi

288 books.
8 books in 5 minutes.
60 minutes in one hour.
5 * 12 = 60
60 minutes * 3 = 3 hours
8 * 12 = 96
96 books in one hour.
96 * 3 = 288
288 books in three hours.

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2 years ago

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How can you estimate 1,506÷2 so that it is close to the actual answer of 753?

Makovka662 [10]

You would round 1506 to 1500 and and do not round 2. So then you woulddo 1500 divided by 2 to get 750!

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2 years ago

Read 2 more answers

Please help me if you can

Mekhanik [1.2K]

Answer:

1)-9x^5 + 7x^2 + 3 2) -2x2 • (x - 7) 3) -23v^2 + 7v + 4 4) is just 7x

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2 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

2 years ago