A - side of small map
a+6 - side of larger map
(a+6)² - area of larger map
(a+6)² = 121
(a+6)² = 11² |√(...)
a+6 = 11 |-6
a+6-6 = 11-6
<span>a = 5
</span>Answer A. <span>5 inches by 5 inches</span>
Answer:
y=−2x−9
Step-by-step explanation:
Answer:
its a glitch
Step-by-step explanation:
Answer:
Step-by-step explanation:
The probability of getting one even and one odd is the sum of probability of getting even and of getting odd and even. Since there are 3 even or odd on each die out of six numbers you have
P(EO)(3/6)(3/6)=1/4 and P(OE)=(3/6)(3/6)=1/4 so 1/4+1/4= 1/2
or you can find the probability of getting two odds or two evens and subtract that from 1
P(EE)=(3/6)(3/6)=1/4, P(OO)=(3/6)(3/6)=1/4
P(EO)=1-1/4-1/4=1/2
The classifications of the functions are
- A vertical stretch --- p(x) = 4f(x)
- A vertical compression --- g(x) = 0.65f(x)
- A horizontal stretch --- k(x) = f(0.5x)
- A horizontal compression --- h(x) = f(14x)
<h3>How to classify each function accordingly?</h3>
The categories of the functions are given as
- A vertical stretch
- A vertical compression
- A horizontal stretch
- A horizontal compression
The general rules of the above definitions are:
- A vertical stretch --- g(x) = a f(x) if |a| > 1
- A vertical compression --- g(x) = a f(x) if 0 < |a| < 1
- A horizontal stretch --- g(x) = f(bx) if 0 < |b| < 1
- A horizontal compression --- g(x) = f(bx) if |b| > 1
Using the above rules and highlights, we have the classifications of the functions to be
- A vertical stretch --- p(x) = 4f(x)
- A vertical compression --- g(x) = 0.65f(x)
- A horizontal stretch --- k(x) = f(0.5x)
- A horizontal compression --- h(x) = f(14x)
Read more about transformation at
brainly.com/question/1548871
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