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2 years ago

Classify the triangle as acute, right, or obtuse and classify it as equilateral, isoceles, or scalene.

Mathematics

2 answers:

julia-pushkina [17]2 years ago

8 0

Answer:

Step-by-step explanation:

An obtuse triangle has ONLY one greater-than-180 angle measure, so this fact rules out B. An equilateral triangle has 3 exact same sides and an isosceles triangle has 2 equal side lengths. Since the "tally marks" is all different, this mean it's neither equilateral nor isosceles, so this fact also rules out A and D. Since a scalene triangle has no equal side length, and the only option available is C. SO THE FINAL ANSWER IS C

Vsevolod [243]2 years ago

5 0

I think it’s the one with acute because it’s less than 90 degrees

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1) Solve ;

Rashid [163]

Answer:

x ≤ 7/2

Step-by-step explanation:

Expand brackets: 3x - 15 ≤ x - 8

Subtract x from both sides: 2x - 15 ≤ -8

Add 15 to both sides: 2x ≤ 7

Divide both sides by 2: x ≤ 7/2

3 0

2 years ago

Read 2 more answers

Hector is flying a kite. He has let out 86 feet

Kaylis [27]

Answer:

Hector's kite is 61.84 feet from the ground.

Step-by-step explanation:

The angle of elevation of the kite is 42°15’30” when converted to decimals, it is $42.258333^{0}$ ≅ $42.26^{0}$

Let the height of the kite to the horizontal of angle of elevation be represented as x. Applying the trigonometric function to the sketch of Hector's kite,

Sin θ = $\frac{opposite}{hypotenus}$

Sin $42.26^{0}$ = $\frac{x}{86}$

⇒ x = 86 x Sin $42.26^{0}$

= 86 x 0.6725

= 57.835

x ≅ 57.84 feet

The height of Hector's kite from the ground = x + 4

= 57.84 + 4

= 61.84 feet

7 0

3 years ago

At the Boston Marathon there were 20,000 runners who finished the race. Each runner ran a distance of 26 miles. If you add toget

andre [41]

Answer:

the runners have gone 20.8 times around the globe all together

Step-by-step explanation:

At the Bostom Marathon, we have a total of

N = 20,000 runners

We know that each runner ran a distance of

d = 26 miles

Therefore, the total distance travelled by all runners combined is:

$D=Nd=(20,000)(26)=5.2\cdot 10^5 mi$

We know that the circumference of the Earth is

$c=2.5\cdot 10^4 mi$

Here we want to find how many times around the globe would the marathon runners have gone. This can be found by calculating the following ratio:

$n=\frac{D}{c}$

And substituting the values, we find:

$n=\frac{5.2\cdot 10^5}{2.5\cdot 10^4}=20.8$

So, the runners have gone 20.8 times around the globe all together.

4 0

3 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

What types of solutions does 6x^2 - 20x + 1 have?

elena55 [62]

Answer:

2 real solutions

Step-by-step explanation:

We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:

b² - 4ac

If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.

Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:

b² - 4ac

(-20)² - 4 * 6 * 1 = 400 - 24 = 376

Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.

<em>~ an aesthetics lover</em>

3 0

3 years ago