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3 years ago

The measures of the angles of a triangle are shown in the figure below. Solve for x.

Mathematics

1 answer:

Daniel [21]3 years ago

6 0

Answer: x = 11 degrees

Step by step explanation:
Triangle rule states that all angles of a trundle must equal to 180 degrees. So we have two known angles, therefore we can subtract them from the 180 degree total.
180-109-43
=28
Now to solve for x we have the equation
4x-16=28
We add 16 to both sides to isolate x
4x=44
Now we divide each side by 4 to get rid of the coefficient
x=11

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Find the product: -3*(-12)

7nadin3 [17]

36. (-)•(-)=(+) 3•12=36 therefore the answer is 36

5 0

3 years ago

Read 2 more answers

Evaluate the expression using the values in the left column to complete the table

egoroff_w [7]

Answer/Step-by-step explanation:

A. $4b + 3b$

Using b = 2

$4(2) + 3(2) = 8 + 6 = 14$

Using b = 4

$4(4) + 3(4) = 16 + 12 = 28$

Using b = 6

$4(6) + 3(6) = 24 + 18 = 42$

Using b = 8

$4(8) + 3(8) = 32 + 24 = 56$

Using b = 10

$4(10) + 3(10) = 40 + 30 = 70$

Using b = 2

$4(12) + 3(12) = 48 + 36 = 84$

B. $3(x - 2)$

Using x = 3

$3(3 - 2) = 3(1) = 3$

Using x = 5

$3(5 - 2) = 3(3) = 9$

Using x = 7

$3(7 - 2) = 3(5) = 15$

Using x = 9

$3(9 - 2) = 3(7) = 21$

Using x = 11

$3(11 - 2) = 3(9) = 27$

Using x = 13

$3(13 - 2) = 3(11) = 33$

C. $5(4a - 1)$

Using a = 2

$5(4(2) - 1) = 5(8 - 1) = 5(7) = 35$

Using a = 3

$5(4(3) - 1) = 5(12 - 1) = 5(11) = 55$

Using a = 4

$5(4(4) - 1) = 5(16 - 1) = 5(15) = 75$

Using a = 5

$5(4(5) - 1) = 5(20 - 1) = 5(19) = 95$

Using a = 6

$5(4(6) - 1) = 5(24 - 1) = 5(23) = 115$

Using a = 7

$5(4(7) - 1) = 5(28 - 1) = 5(27) = 135$

D. $6y - 2y + 4$

Using y = 2

$6(2) - 2(2) + 4 = 12 - 4 + 4 = 12$

Using y = 3

$6(3) - 2(3) + 4 = 18 - 6 + 4 = 18 - 2 = 16$

Using y = 5

$6(5) - 2(5) + 4 = 30 - 10 + 4 = 30 - 6 = 24$

Using y = 10

$6(10) - 2(10) + 4 = 60 - 20 + 4 = 60 - 16 = 44$

Using y = 8

$6(8) - 2(8) + 4 = 48 - 16 + 4 = 48 - 12 = 36$

Using y = 6

$6(6) - 2(6) + 4 = 36 - 12 + 4 = 36 - 8 = 28$

7 0

3 years ago

find a model for simple harmonic motion if the position at t = 0 is 6, the amplitude is 6 centimeters, and the period is 4 secon

Y_Kistochka [10]

Answer:

y = 6cos(π/2)t

Step-by-step explanation:

Given in the question that,

in a simple harmonic motion in which

at t=0 the amplitude is 6 cm, so the simple harmonic motion will be in cosine form. As in case of sine function, the value of function at t=0 is 0.

And,

the period is given as 4 sec

We know that model equation of simple harmonic motion is

here,

a is amplitude

c is Horizontal shift or phase shift

d is Vertical shift

Period = 2 π / b

4 = 2π / b

b = 2π/4

b = 1π/2

<h3>Step2</h3><h3>Find shifts</h3>

Horizontal and vertical shifts are 0.

So, c = 0 and d = 0

<h3>Step3</h3><h3>Plug those values in the above in the equation</h3>

y = 6cos π/2 (t + 0) + 0

y = 6cos(π/2)t

4 0

4 years ago

How to graph y=2^x+3

Mashcka [7]

Answer:

see below

Step-by-step explanation:

It is usually convenient to choose small exponents when graphing an exponential function. You can get a reasonable idea of the shape of the curve using x-values with a magnitude of 3 or less.

The exponential term 2^x has a horizontal asymptote of y=0 for large negative values of x. Adding 3 to that term shifts the horizontal asymptote up to y=3. Of course, everything else is shifted up the same amount.

You know that ...

2^-1 = 1/2

2^0 = 1

2^1 = 2

2^2 = 4

Adding 3 to these values will give you points on the graph for x=-1 to 2.