What is the difference between similar and congruent

Mathematics

2 answers:

lakkis [162]3 years ago

8 0

Similar look the same but might not have same sie congruent is exactly the same

sattari [20]3 years ago

6 0

Similar shapes can be different sizes, but look the same. Congruent shapes are the exact same as eachother.

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5. Each pair of polygons below are similar. Show your work! Solve for x:

Stella [2.4K]

Since they are similar, you need to find the ratio of similarity (I made up the term, there is probably a correct one that I can’t remember).

If you divide 16/40, you’ll find that that ratio is 2.5. So then you just multiply 16 x 2.5. You’ll get 18.

18 is the length of the top of the trapezoid.

You set 18=2x+4 and solve it algebraically. Subtract 4 from both sides.

14=2x
Divide by 2 and x=7

(You can also check that the ratio is right by 16/18 is the same decimal value as 40/45. You’ll get .88888...)

4 0

3 years ago

Evaluate the following expression. cos135°

Mrac [35]

Negative 1 / Square Root of 2

4 0

3 years ago

Which products result in a difference of squares or a perfect square trinomial? Check all that apply.

yawa3891 [41]

(5x + 3)(5x – 3)
(7x + 4)(7x + 4)
(x – 9)(x – 9)
(–3x – 6)(–3x + 6)

6 0

3 years ago

Read 2 more answers

The Zurn family ate at a restaurant that was having a 15% discount on that day.Their meal costs $64.75,and they left a 20% tip.W

My name is Ann [436]

Answer:C

Step-by-step explanation:

4 0

3 years ago

How do you find the absolute extrema of this function?<br> f(x) = x+3x^(2/3); Interval is [-10,1]

Step2247 [10]

$\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}} \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x}
\\\\\\
(-2)^3=x\implies \boxed{-8=x}\\\\
-------------------------------\\\\
0=\sqrt[3]{x}\implies \boxed{0=x}$

now, f(0) = 0, and f(-8) is an imaginary value or no real value.

now, f(-10) will also give us an imaginary value

and f(1) = 4

so, doing a first-derivative test on 0, is imaginary to the left and positive on the right, and before and after 1, is positive as well, so f(x) is going up on those intervals.

however, f(0) is 0 and f(1) is higher up, so the absolute maximum will have to be f(1), and we can use f(0) as a minimum, and since it's the only one, the absolute minimum.

the other two, the endpoint of -10 and the critical point of -8, do not yield any values for f(x).

8 0

3 years ago