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4 years ago

Classify the triangle with angles measuring 40°, 40°, and 100°.

Mathematics

2 answers:

bagirrra123 [75]4 years ago

8 0

40 and 40 are acute and 100 is obtuse

ASHA 777 [7]4 years ago

6 0

Idk if you have a option D because that would be an iso<span>sceles </span>

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Answer correct and fast

inna [77]

Hello there I hope you are having an great Day :)

1) 32780ml = 32.78

2) 37 1/2% of 24 = 9

Hopefully that helps you :)

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3 years ago

an isosceles triangle has a vertex angle of 21.21 degrees. two sides of the triangle are each 17.91 ft long. whats the area of t

lubasha [3.4K]

Answer:

The area of triangle is $A=58.02\ ft^{2}$

Step-by-step explanation:

Let

B -----> the measure of the vertex angle of the isosceles triangle ABC

a and c -----> the two congruent sides of the isosceles triangle ABC

we know that

The area of a triangle applying the law of sines is equal to

$A=\frac{1}{2}(a)(c)sin(B)$

we have

$a=c=17.91\ ft$

$B=21.21\°$

substitute in the formula

$A=\frac{1}{2}(17.91)(17.91)sin(21.21\°)$

$A=58.02\ ft^{2}$

7 0

4 years ago

Factor the polynomial-x^3-4x^2-5x

Gnom [1K]

Answer:

(-x)(x^2 + 4x + 5)

Step-by-step explanation:

You have to factor out the -x from the rest of the expression.

5 0

3 years ago

39*10*100*1,000*0+37???? HELPPP

FromTheMoon [43]

Answer:

Step-by-step explanation:

Anything multiplied by 0 is 0 then you just add the 37

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3 years ago

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2−x2. What are the dimensions of su

Paul [167]

Answer:

Step-by-step explanation:

Given that a rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

$y=2-x^2$

the parabola is open down with vertex at (0,2)

We can find that the rectangle also will be symmetrical about y axis.

Let the vertices on x axis by (p,0) and (-p,0)

Then other two vertices would be (p,2-p^2) (-p,2-p^2) because the vertices lie on the parabola and satisfy the parabola equation

Now width = $2-p^2$

Area = l*w = $2(2p-p^3)$

Use derivative test

I derivative = $2(2-3p^2)$

II derivative = $-12p$

Equate I derivative to 0 and consider positive value only since we want maximum

p = $\sqrt{\frac{2}{3} }$

Thus width= $\sqrt{\frac{2}{3} }$

Length = $2\sqrt{\frac{2}{3} }$

Width = $2-2/3 = 4/3$

3 0

3 years ago