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3 years ago

What the answer question

Mathematics

1 answer:

algol [13]3 years ago

5 0

Answer:

∠WUV

Step-by-step explanation:

Start from the top, the meeting point of the rays is the angle U

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Y= 7x - 19 y= 3x - 7 Solve for x and then use X to solve for y. (find x and y)

stepladder [879]

Answer:

(2,3)

Step-by-step explanation:

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2 years ago

Read 2 more answers

SOMEONE PUT AN ANSWER SO IT DOESNT GET DELETED!!!

Nadusha1986 [10]

Answer:

The height of the cylinder is 4x units.

Step-by-step explanation:

volume of a cylinder = nr^2h

n = 22/7

r = radius = x/2

Let illustrate with an example. If the diameter of a cylinder is 6 and the volume is 6^3 = 216. The radius = 6/2 = 3

n3^2 x h = 216

h = 216/9 = 24

so the height is 4 times the diameter

Area of the base = nr^2

1/2^2 x x = 1/4 pi x^2

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2 years ago

Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0

Masja [62]

|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

--------------------------------------------------------------------------------

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

--------------------------------------------------------------------------------

$h(x)=\dfrac{1}{3}|x-6|+4$

Put the values of x to the equation of the function h(x):

$x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6$

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2 years ago

What number can I put in the denominator? 1 8 7

Zigmanuir [339]

Answer:

The LCD denominator is 1

Step-by-step explanation:

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3 years ago

Which simplifications of the powers of i are correct? There may be more than one correct answer.

fredd [130]

$\bf i^2=-1\qquad\qquad i^3=-i\qquad \qquad i^4=1 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ i^{22}\implies i^{(4\cdot 5)+2}\implies i^{4\cdot 5}i^2\implies (i^4)^5 i^2\implies 1^5(-1)\implies -1~\dotfill \bigotimes \\\\\\ i^{11}\implies i^{(2\cdot 5)+1}\implies (i^2)^5 i\implies (-1)^5(i)\implies -i~\dotfill \checkmark$

$\bf i^{21}\implies i^{(4\cdot 5)+1}\implies (i^4)^5 i\implies 1^5(i)\implies i~\dotfill \checkmark \\\\\\ i^{12}\implies i^{3\cdot 4}\implies i^3 i^4\implies (-i)(1)\implies -i\dotfill \bigotimes \\\\\\ i^{20}\implies i^{4\cdot 5}\implies (i^4)^5\implies 1~\dotfill \checkmark \\\\\\ i^{26}\implies i^{(4\cdot 6)+2}\implies (i^4)^6 i^2\implies 1^6(-1)\implies -1\dotfill \checkmark \\\\\\ i^{27}\implies i^{(4\cdot 6)+3}\implies (i^4)^6 i^3 \implies 1^6(-i)\implies -i\dotfill \bigotimes$

6 0

3 years ago