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rusak2 [61]
3 years ago
9

Select all the triangles that have an area of 30 square units.

Mathematics
1 answer:
marta [7]3 years ago
6 0

Answer:

A and C

Step-by-step explanation:

Area of ∆ = ½*base*height

Area of figure A:

base = 6

height = 10

Area = ½*6*10 = 3*10 = 30 square units

Area of figure B:

base = 10

height = 8.66

Area = ½*10*8.66 = 5*8.66 = 43.3 square units

Area of figure C:

base = 10

height = 6

Area = ½*10*6 = 5*6 = 30 square units

Area of figure D:

base = 6

height = 8

Area = ½*6*8 = 3*8 = 24 square units

Area of figure E:

base = 3

height = 10

Area = ½*3*10 = 3*5 = 15 square units

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Things to remember the sum of the measures of the interior and exterior angles at each vertex is 180 (supplementary angles) add and equal to 180 3x-5+5x+17= 180 8x+12=180 8x= 168 x=21 substitute in the algebraic expression for exterior angle 3(21) -5= 63-5=58 58 degrees is the exterior angle hope this helps
8 0
3 years ago
What other information do you need in order to prove the triangles congruent using the SAS congruence postulate?
ziro4ka [17]

Answer:

I think D

Step-by-step explanation:

I think that the answer is D, since the angles are not visibly shown to be equal. If you knew answer D, you would then be able to see that segments BC and CD are congruent. I am not 100% sure, however.

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3 years ago
A food packet is dropped from a helicopter and is modeled by the function f(x) = −15x2 + 6000. The graph below shows the height
melisa1 [442]

General Idea:

Domain of a function means the values of x which will give a DEFINED output for the function.

Applying the concept:

Given that the x represent the time in seconds, f(x) represent the height of food packet.

Time cannot be a negative value, so

x\geq 0

The height of the food packet cannot be a negative value, so

f(x)\geq 0

We need to replace -15x^2+6000 for f(x) in the above inequality to find the domain.

-15x^2+6000\geq 0 \; \;  [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0

The possible solutions of the above inequality are given by the intervals (-\infty , -2], [-2,2], [2,\infty ). We need to pick test point from each possible solution interval and check whether that test point make the inequality (x+200)(x-200)\leq 0 true. Only the test point from the solution interval [-200, 200] make the inequality true.

The values of x which will make the above inequality TRUE is -200\leq x\leq 200

But we already know x should be positive, because time cannot be negative.

Conclusion:

Domain of the given function is 0\leq x\leq 200

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Answer:

Step-by-step explanation:

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Use the arc length formula to find the length of the curve y = 2 − x2 , 0 ≤ x ≤ 1. Check your answer by noting that the curve is
Lina20 [59]

By using the arc length formula, we will see that the length of the curve is L = 1.48

<h3>How to use the arc length formula?</h3>

Here we have the curve:

y = 2 - x^2   with 0 ≤ x ≤ 1

And we want to find the length of the curve.

The arc length formula for a curve y in the interval [x₁, x₂] is given by:

L  =\int\limits^{x_2}_{x_1} {\sqrt{1 + \frac{dy}{dx}^2} } } \, dx

For our curve, we have:

dy/dx = -2x

And the interval is [0, 1]

Replacing that we get:

L  =\int\limits^{1}_{0} {\sqrt{1 + (-2x)^2} } } \, dx\\\\L  \int\limits^{1}_{0} {\sqrt{1 + 4x^2} } } \, dx\\

This integral is not trivial, using a table you can see that this is equal to:

L = (\frac{Arsinh(2x)}{4}  + \frac{x*\sqrt{4x^2 + 1} }{2})

evaluated from x = 1 to x = 0, when we do that, we will get:

L = \frac{Arsinh(2) + 2*\sqrt{5} }{4}  = 1.48

That is the length of the curve.

If you want to learn more about curve length, you can read:

brainly.com/question/2005046

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