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3 years ago

What is the area of the trapezoid with height 13 units?

Mathematics

2 answers:

Verizon [17]3 years ago

7 0

The equation for Area of Trapezoid is:

$Atrapezoid = \frac{h(b1+b2)}{2}$

First, replace the variables with their values:

<em>h</em>=13

Now, your equation should look like this:

$Atrapezoid = \frac{13(15+29)}{2}$

Now, solve:

$15+29=44$

↓

$\frac{13(44)}{2}$

↓

$44*13=572$

↓

$\frac{572}{2}=286$

The area of the trapezoid is 286.

Hope this helps! Please let me know if you need more help, or if you think my answer is incorrect. Brainliest would be MUCH appreciated. Have a great day!

Stay Brainy!

- $Kallmekris$

zhannawk [14.2K]3 years ago

5 0

Answer:

286

Step-by-step explanation:

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The points S, T, U and V all lie on the same line segment, in that order, such that the

Evgesh-ka [11]

<h3>Answer: TV = 10</h3>

==============================================

Explanation

Let x be some positive real number such that

ST = 6x
TU = 2x
UV = 3x

leading to the ratio

ST:TU:UV = 6x:2x:3x = 6:2:3

Note how I effectively started with 6:2:3 and multiplied all parts of the ratio by x to get 6x:2x:3x

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

The parts of the ratio add up to the longest segment SV

ST+TU+UV = SV

6x+2x+3x = 22

11x = 22

x = 22/11

x = 2

Therefore,

ST = 6x = 6*2 = 12
TU = 2x = 2*2 = 4
UV = 3x = 3*2 = 6

Note how ST+TU+UV = 12+4+6 = 22, which is the length of SV. This helps confirm we have the correct x value.

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

The last step is to find the length of TV

TV = TU+UV

TV = 4+6

TV = 10

Or you could do

ST + TV = SV

12+TV = 22

TV = 22-12

TV = 10

8 0

3 years ago

Use differentiation to show that the given function is a solution of the equation for all values of the constants. (Enter your a

Katena32 [7]

Answer with Step-by-step explanation:

We are given that

$x''+x=2e^t$

Function: $x=C_1sint+C_2cost+e^t$

We have to show that given function is a solution of the equation for all values of the constants.

If given function is solution of DE then it satisfied the given DE.

Differentiate function w.r.t.t

$x'=C_1cost-C_2sint+e^t$

Again differentiate w.r.t. t

$x''=-C_1sint-C_2cost+e^t$

Substitute the values in the given DE

$-C_1sint-C_2cost+e^t+C_1sint+C_2cost+e^t=2e^t$

LHS=RHS

Given function satisfied the given DE.Therefore, it is solution of given DE for all values of the constants.

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3 years ago

Determine whether lines BT and MV are parallel, perpendicular, or neither. You must show all of your work to earn full credit.

Dafna1 [17]

<h3>Answer:</h3>

BT ⊥ MV

<h3>Explanation:</h3>

The direction vector BT is ...

... T - B = (5, 12) - (1, -4) = (4, 16)

The direction vector MV is ...

... V - M = (-4, 2) - (-8, 3) = (4, -1)

One is not a multiple of the other, so the lines are <em>not parallel</em>.

The dot-product of these direction vectors is

... (4, 16)•(4, -1) = 4·4 + 16·(-1) = 0

When the dot-product is zero, the vectors are perpendicular.

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Answer:

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Step-by-step explanation:

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thank you.

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The equation that best models this situation would be: y=40+30x

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