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3 years ago

What is the perimeter of the parallelogram? Please help! Will mark brainlyest.

Mathematics

2 answers:

Anon25 [30]3 years ago

8 0

Step-by-step explanation:

58 maybe coz I got 58 so??

Luba_88 [7]3 years ago

3 0

Answer:

Step-by-step explanation:

B=base

A=side

X=2(a+b)=2·(13+16)=58

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3 years ago

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3 years ago

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zaharov [31]

Answer:

7 ÷ 5

Step-by-step explanation:

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4 years ago

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What is the volume of the composite figure shown? Use 3.14 for π.

Ivenika [448]

Answer:

25.12

Step-by-step explanation:

the formula of a cone is pi r^2 * h/3

r = radius and h = height

the radius is 2 and the height is 3

and the question says use 3.14 instead of pi so

we have

3.14 * 2^2 * 6/3

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6/3 = 2

3.14 * 4 * 2

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and 8 * 3.14 is 25.12

7 0

3 years ago

Read 2 more answers

The function f(x) = x is translated using the rule (x, y) - (x - 6, y + 9) to create A(x).

anyanavicka [17]

It seems the correct function you need to mention is f(x) = sqrt x

so, I am assuming $f(x) = \sqrt{x}$ and will solve the question based on it.

Answer:

Check the explanation

Step-by-step explanation:

Given the function

$f(x) = \sqrt{x}$

As the function is translated according to the rule

(x, y) → (x - 6, y + 9)

Translation of the function $f(x) = \sqrt{x}$ 6 units to the left will bring the function

$g(x) = \sqrt{x+6}$

Translation of the function $g(x) = \sqrt{x+6}$ 9 units up will bring the function

$A\left(x\right)=\sqrt{x+6}+9$

Determining the range of $A\left(x\right)=\sqrt{x+6}+9$

As we know that the range is the set of dependent values for which the function is defined.

$\mathrm{The\:range\:of\:an\:radical\:function\:of\:the\:form}\:c\sqrt{ax+b}+k\:\mathrm{is}\:\:f\left(x\right)\ge \:k$

$k=9$

$f\left(x\right)\ge \:9$

$\mathrm{Range\:of\:}\sqrt{x+6}+9:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:9\:\\ \:\mathrm{Interval\:Notation:}&\:[9,\:\infty \:)\end{bmatrix}$

Therefore, the expression describes the range of A(x) will be:

$f\left(x\right)\ge \:\:9\:or\:y\:\ge \:\:\:9$

3 0

3 years ago