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2 years ago

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Rhombus ABCD ~Rhombus EFGH. The scale factor is 3. Angle A = 80° , Angle B = 70° Angle C = 110°, and Angle D = 100°.What are the

measures of Angle E, Angle F, Angle G, and Angle H Group of answer choices

1 answer:

sukhopar [10]2 years ago

7 0

Answer:

d 100

Step-by-step explanation:

You might be interested in

Show why x−3 is a factor of m(x)=x3−x2−5x−3. Justify your answer. You can show this using remainder theorem, long division, synt

Marat540 [252]

The factors of a polynomial function are the zeros of the function

It is true that x - 3 is a factor of m(x) = x^3 - x^2 - 5x - 3

<h3>How to show why the x - 3 is a factor</h3>

The function is given as:

m(x) = x^3 - x^2 - 5x - 3

The factor is given as:

x - 3

Set the factor to 0

x - 3 = 0

Solve for x

x = 3

Substitute 3 for x in the function

m(3) = 3^3 - 3^2 - 5(3) - 3

Evaluate

m(3) =0

Since the value of m(3) is 0, then x - 3 is a factor of m(x) = x^3 - x^2 - 5x - 3

Read more about factors at:

brainly.com/question/11579257

7 0

2 years ago

Show work please do

quester [9]

Here is your answer! I hope this helps.

6 0

3 years ago

A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate

Lyrx [107]

Answer:

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

$\frac{dV}{dt}=4 \ m^3/s$

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is $V=\pi r^2 h$

$=( \pi \times 2^2\times h)\ m^3$

$\therefore V= 4\pi h$

Differentiating with respect to t.

$\frac{dV}{dt}=4\pi \frac{dh}{dt}$

Putting the value $\frac{dV}{dt}$

$\Rightarrow 4\pi \frac{dh}{dt}=4$

$\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}$

The rate change of height does not depend on the height.

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

3 0

3 years ago

Need help solving these 3 equations, please help.

pantera1 [17]

1.) r=-1/3
2.) x=-33/29
3.) x=-4/5 x=2/5

3 0

3 years ago

(01.03. 106, 1.08 HC)

zlopas [31]

Answer:

A) see attached for a graph. Range: (-∞, 7]

B) asymptotes: x = 1, y = -2, y = -1

C) (x → -∞, y → -2), (x → ∞, y → -1)

Step-by-step explanation:

<h3>Part A</h3>

A graphing calculator is useful for graphing the function. We note that the part for x > 1 can be simplified:

$\dfrac{-x^2+x+2}{x^2-3x+2}=-\dfrac{(x-2)(x+1)}{(x-2)(x-1)}=-\dfrac{x+1}{x-1}\quad x\ne 2$

This has a vertical asymptote at x=1, and a hole at x=2.

The function for x ≤ 1 is an ordinary exponential function, shifted left 1 unit and down 2 units. Its maximum value of 3^-2 = 7 is found at x=1.

The graph is attached.

The range of the function is (-∞, 7].

__

<h3>Part B</h3>

As we mentioned in Part A, there is a vertical asymptote at x = 1. This is where the denominator (x-1) is zero.

The exponential function has a horizontal asymptote of y = -2; the rational function has a horizontal asymptote of y = (-x/x) = -1. The horizontal asymptote of the exponential would ordinarily be y=0, but this function has been translated down 2 units.

__

<h3>Part C</h3>

The end behavior is defined by the horizontal asymptotes:

for x → -∞, y → -2

for x → ∞, y → -1

7 0

2 years ago