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2 years ago

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riangle ABC is an obtuse triangle with the obtuse angle at vertex B. Angle A must be less than 90°. greater than 90°. congruent

to angle B. congruent to angle C.

1 answer:

Softa [21]2 years ago

5 0

Angle A must be less than 90<span>°</span>

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Alik [6]

Multiply 14 over 7 to get the number as one fraction. after adding three to that product, you get 101/7. Multiply that by 6/1 to get 606/7, then divide 606 by 7 to get 86 4/7 hours every 6 months.

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3 years ago

Read 2 more answers

The volumes of two similar solids are 125 cm^3 and 1,000 cm^3. The surface area of the smaller solid is 150 cm^2. What is the su

Flura [38]

<u>Find scale Factor:</u>

<u>
</u>

$(\cfrac{length_1}{length_2})^3 = \cfrac{volume_1}{volume_2}$

$(\cfrac{length_1}{length_2})^3 = (\cfrac{125}{1000} )$

$\cfrac{length_1}{length_2} = \sqrt[3]{\cfrac{125}{1000} }$

$\cfrac{length_1}{length_2} = \cfrac{5}{10} = \cfrac{1}{2}$

<u>Find Surface area of the larger solid:
</u>

$(\cfrac{length_1}{length_2} )^ 2=\cfrac{area_1}{area_2}$

$(\cfrac{1}{2} )^ 2=\cfrac{150}{area_2}$

$\cfrac{1}{4} =\cfrac{150}{area_2}$

$area_2 = 150 \times 4 = 600 \text{ cm}^2$

Answer: 600 cm²

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3 years ago

Which best describes the relationship between the successive terms in the sequence shown?

tensa zangetsu [6.8K]

This is a geometric sequence where each number is being multiplied by -2

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2 years ago

Instructions: Find m

Maurinko [17]

Check the picture below.

we could also look at it from the inscribed angle theorem, bearing in mind that the intercepted arc is 180°.

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1 year ago

Write the vector v in the form aiplusbj, given its magnitude left norm Bold v right norm and the angle alpha it makes with the

photoshop1234 [79]

Answer:

$6\sqrt{3}i-6j$

Step-by-step explanation:

Given:

The magnitude, v=12

The angle $\alpha$ it makes with the positive x-axis = $150^0$

Now for a vector (x,y)

$\alpha =tan^{-1}(\frac{y}{x} )$ ; and

$v=\sqrt{x^2+y^2}$

Therefore:

$150 =tan^{-1}(\frac{y}{x} )\\\frac{y}{x} =tan 150\\\frac{y}{x}=-\frac{\sqrt{3} }{3}$

$y=-\frac{x \sqrt{3}}{3}$

Similarly:

$v=\sqrt{x^2+y^2}\\12=\sqrt{x^2+y^2}\\12^2=x^2+y^2\\144=x^2+y^2$

$144=x^2+(-\frac{x \sqrt{3}}{3} )^2\\144=x^2+\frac{3x^2}{9} \\144=\frac{4x^2}{3}\\4x^2=144 X 3=432\\x^2=108\\x=6\sqrt{3}$

Recall that:

$y=-\frac{x \sqrt{3}}{3}\\y=-\frac{6\sqrt{3} \sqrt{3}}{3}=-6$

Therefore: $(x,y)=(6\sqrt{3}, -6)$

The vector in the form ai+bj is therefore:

$6\sqrt{3}i-6j$

6 0

3 years ago