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3 years ago

Can anyone help me with this im having some difficulty

Mathematics

2 answers:

Whitepunk [10]3 years ago

7 0

Answer:

option C

Step-by-step explanation:

$(a+b)(a+b) = a^2 + 2ab + b^2$

amm18123 years ago

4 0

Answer:

Step-by-step explanation:

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I NEED SOMEONE FROM RSM 7TH GRADE TO COME AND HELP ME ASAP PLZ!!!!!!!!!!!!

Aneli [31]

Answer:

I don't see any problem or I would help u out

4 0

2 years ago

Which graph has a slope of ? A coordinate plane with a straight line. The line starts at (negative 5, negative 4) and passes thr

VikaD [51]

Answer:

slope of the first line: 1

slope of the second line: 0.778

slope of the third line: 0.375

slope of the fourth line: 1.25

Step-by-step explanation:

Given two points (x1, y1) and (x2, y2), the slope of a line is computed as follows:

slope = (y2 - y1)/(x2 - x1)

Therefore,

slope of the first line: [5 - (-4)]/[4 - (-5)] = 1

slope of the second line: [5 - (-2)]/[4 - (-5)] = 0.778

slope of the third line: [2 - (-1)]/[3 - (-5)] = 0.375

slope of the fourth line: [5 - (-5)]/[4 - (-4)] = 1.25

5 0

3 years ago

Read 2 more answers

Please help me What is the quotient when 2/3 is divided by 5/6

Margaret [11]

Answer:

You on? I’m bored! You- you on? I’m bored

Step-by-step explanation:

4 0

3 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

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