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2 years ago

7

Two angles are supplementary. The smaller angle is 35 degrees less than the larger angle. What is the measurement of the larger

angle?

1 answer:

svetlana [45]2 years ago

5 0

Answer:

Step-by-step explanation:

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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

Complete the point-slope equation of the line through (-1,6)(−1,6)left parenthesis, minus, 1, comma, 6, right parenthesis and (1

velikii [3]

Answer:

$y-6 = -\frac{1}{2} (x+1)$

Step-by-step explanation:

For this case we have two points given (-1,6) and (1,5)

And we want to complete the point slope equation of the line:

$y-6 = m(x-x_o)$

We need to find the slope and we can use the following formula:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

And replacing the info we got:

$m= \frac{5-6}{1-(-1)}=-\frac{1}{2}$

And then the equation would be given by:

$y-6 = -\frac{1}{2} (x- (-1))$

And our final answer would be:

$y-6 = -\frac{1}{2} (x+1)$

3 0

2 years ago

gavin deposited $1500 into his savings account that is compounded quarterly at an interest rate of 1.5%. How munch money will ga

creativ13 [48]

<u>Answer:</u>

The money will gavin have after 5 years is 1616.59$

<u>Explanation:</u>

We know that compound interest is given by

$A=P\left(1+\frac{r}{n}\right)^{n t}$

Where A = final amount

P = Principal amount = $1500 (given)

r = interest rate = 1.5% = 0.015

n = no. of times interest applied per time period = given quarterly = 4

t = time period = 5 years

So,

$A=1500\left(1+\frac{0.015}{4}\right)^{4 \times 5}$

$1500\left(1+\frac{0.015}{4}\right)^{20}$

= 1616.59$ which is the money will gavin have after 5 years

4 0

3 years ago

What are the relative minimum and relative maximum values over the interval [−4,4] for the function shown in the graph?

kobusy [5.1K]

Answer:

Relative minimum : -36

Relative maximum : 64

The rate of change is 336 greater

Step-by-step explanation:

Relative minimum are the minimum values in the interval

Looking at the graph, we find the lowest point in the interval

Relative minimum : (-3, -36) and (3,-36) y value -36

Looking at the graph, we find the highest point in the interval

Relative maximum : (0,64) y value 64

Average rate of change = f(x2) - f(x1)

---------------

x2 - x1

f(7) - f(5) 1469 - 549 920

------------- = --------------- = ------- = 460

7-5 7-5 2

f(4) - f(2) 287 - 39 248

------------- = --------------- = ------- = 124

4-2 4-2 2

We need to subtract

460-124

336

6 0

3 years ago

Read 2 more answers

during track practice Pedro runs a lap in 3 minutes. David runs the same lap in 4 minutes. they both pass point S at the same ti

garri49 [273]

Im not sure this is right but i think they will meet during 12 minutes

8 0

3 years ago