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2 years ago

6

Y=a|x-h|+k please help

1 answer:

RSB [31]2 years ago

7 0

Answer:

i can't answer this but i can tell you this

The general form of an absolute value function is f(x)=a|x-h|+k. From this form, we can draw graphs.

y = a(x – h)2 + k, where (h, k) is the vertex. ... In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "x – h" is squared, so its value is always zero or greater; being squared, it can never be negative.

Step-by-step explanation:

it is not the answer but i hope it helps:)

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Molodets [167]

Answer:

The supplement is 94

Step-by-step explanation:

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Let x be the supplementary angle

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x+86-86 =180-86

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Which figure has the same area as the parallelogram shown below? A. A triangle with a base of 4 mm and a height of 20 mm B. A tr

Arisa [49]

<h2>Answer: A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>

The parallelogram in the figure has an area of $80mm^{2}$ , according to the following formula, which works for all rectangles and parallelograms:

$A_{parallelogram}=(b)(h)$ (1)

Where $b$ is the base and $h$ is the height

The<u> area of a triangle</u> is given by the following formula:

$A_{triangle}=\frac{1}{2}(b)(h)$ (2)

So, for option A:

$A_{triangle}=\frac{1}{2}(4mm)(20mm)=40mm^{2} \neq 80mm^{2}$

Now, the <u>area of a trapezoid </u>is:

$A_{trapezoid}=\frac{1}{2}(b_{1}+ b_{2})(h)$ (3)

For option B:

$A_{trapezoid}=\frac{1}{2}(15mm+25mm)(2 mm)=40mm^{2} \neq 80mm^{2}$

For option C:

$A_{trapezoid}=\frac{1}{2}(6mm+14mm)(8 mm)=80mm^{2}$ >>>>This is the correct option!

For option D:

$A_{rectangle}=(30mm)(8mm)=240mm^{2} \neq 80mm^{2}$

<h2>Therefore the correct option is C</h2>

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3 years ago

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