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3 years ago

If 3 triangles have the same angles but not the same length, does that make them congruent?

1 answer:

galina1969 [7]3 years ago

4 0

When two triangles are congruent they will have exactly the same three sides and exactly the same three angles. The equal sides and angles may not be in the same position (if there is a turn or a flip), but they are there.

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Log based question, i just need help with parts b and c

Tanya [424]

Answer:

b) There is translation 3 units to the right and 1 unit up

c) The domain is {x I x > 3}

The equation of the asymptote is x = 3

Step-by-step explanation:

* Lets revise the rule of the translation

- If the function f(x) translated horizontally to the right

by h units, then the new function g(x) = f(x - h)

- If the function f(x) translated horizontally to the left

by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up

by k units, then the new function g(x) = f(x) + k

- If the function f(x) translated vertically down

by k units, then the new function g(x) = f(x) – k

* Now lets solve the problem

b) ∵ f(x) = $log_{2} (x-3)+1$

∵ The parent function is $log_{2} x$

∴ x is changed to (x - 3), that means there is a translation 3 units

to the right

∵ We add the parent function by 1, that means there is a translation

1 unit up

* There is translation 3 units to the right and 1 unit up

c) To find the domain of the function, find the values of x which

make the function undefined

∵ $log_{2}(0)$ is undefined

∴ x - 3 can not be 0

∵ x - 3 = 0 ⇒ add 3 to both sides

∴ x = 3

∴ The domain of the function is all real number greater than 3

* The domain is {x I x > 3}

∵ x can not be 3

∴ There is a vertical asymptote, its equation is x = 3

* The equation of the asymptote is x = 3

# Look to the attached graph for more understand for the domain

and the equation of the asymptote

4 0

3 years ago

Please help I don’t understand

irakobra [83]

Answer:

a. 79

Step-by-step explanation:

Use this theorem

$x = \frac{101 + 57}{2 } \\ x = \frac{158}{2} \\ x = 79$

5 0

2 years ago

A quadrilateral has vertices at A(–5, 5), B(1, 8), C(4, 2), D(–2, –2). Use slope to determine if the quadrilateral is a rectangl

lianna [129]

With the properties of rectangle in mind we must perform two verification. Verify to see if opposite sides are parallel and adjacent sides are perpendicular.

We need to determine the slope of each side, using the formula,

$m=\frac{y_2-y_1}{x_2-x_1 }$

Slope of AB

$m_{AB}=\frac{8-5}{1--5}$

$\Rightarrow m_{AB}=\frac{8-5}{1+5}$

$\Rightarrow m_{AB}=\frac{3}{6}$

$\Rightarrow m_{AB}=\frac{1}{2}$

Slope of BC

$m_{BC}=\frac{2-8}{4-1}$

$\Rightarrow m_{BC}=\frac{-6}{3}$

$\Rightarrow m_{BC}=-2$

Slope of CD

$m_{CD}=\frac{-2-2}{-2-4}$

$\Rightarrow m_{CD}=\frac{-2-2}{-2-4}$

$\Rightarrow m_{CD}=\frac{-4}{-6}$

$\Rightarrow m_{CD}=\frac{2}{3}$

Slope of AD

$m_{AD}=\frac{-2-5}{-2--5}$

$\Rightarrow m_{AD}=\frac{-2-5}{-2+5}$

$\Rightarrow m_{AD}=\frac{-7}{3}$

Verify Parallel sides

If the quadrilateral is a rectangle, then opposite sides should have the same slope. But

$m_{AD} = \frac{-7}{3} \neq m_{BC}=-2$

$m_{AB}=\frac{1}{2} \neq m_{CD}=\frac{2}{3}$

Verify Perpendicularity

And also the product of slopes of all sides with a common vertex should be -1. But

$m_{AB} \times m_{BC}=\frac{1}{2} \times -2=-1$

$m_{AB} \times m_{AD}=\frac{1}{2} \times -\frac{7}{3} \neq -1$

$\Rightarrow m_{CD} \times m_{AD}=\frac{2}{3} \times -\frac{7}{3} \neq -1$

$\Rightarrow m_{CD} \times m_{BC}=\frac{2}{3} \times -2 \neq -1$

Since the quadrilateral fails to satisfy all these conditions, the quadrilateral is not a rectangle

6 0

3 years ago

What is 33.3 as a whole number?

Amiraneli [1.4K]

If you mean by rounding, then it is 33

7 0

3 years ago

Idk how to do this ?!?! HELP

NISA [10]

It's asking you what percent of 500 is 430, and you can answer that by first dividing 500 by 100 to find 1%, which is 5, and then dividing 430 by 5, which is 86%, which is your answer. To fill in your sheet, you put the 430 in the left box and the 500 on the right. I hope this helps!

7 0

4 years ago