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3 years ago

Someone please help me ASAP

Mathematics

1 answer:

IgorLugansk [536]3 years ago

6 0

What do you need help with. I don’t see any questions or equations being found but have a great day :)

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A woman at your company saves 10% of her salary.what fraction of her salary doe she save?

elena55 [62]

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3 years ago

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What is 4.8x ten to the power of 8

dangina [55]

480,000,000 (480 million)

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3 years ago

What the common Denominator of 7/8 and 3/10?

ankoles [38]

The common denominator would be 40

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2 years ago

Can someone solve this for me please.

nika2105 [10]

Answer:

The value of PR is 21 ft.

Step-by-step explanation:

We know that:

3(PQ + QT + TP) = PR + RS + SP
PQ = 7 ft.
QT = 4 ft
RS = 12 ft.

Work:

3(PQ + QT + TP) = PR + RS + SP
=> 3(7 + 4 + TP) = PR + 12 + SP
=> 21 + 12 + 3TP = PR + 12 + SP
=> 21 + 3TP = PR + SP
=> 21 + SP = PR + SP [3TP = SP]
=> 21 = PR

Hence, the value of PR is 21 ft.

Hoped this helped.

$BrainiacUser1357$

6 0

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

3 years ago