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3 years ago

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A sideboat cabinet is 90 cm tall, 176 cm wide, and 65 cm deep. What are the dimensions of a scale model built using a scale of 1

:4?

1 answer:

horrorfan [7]3 years ago

8 0

90/4 = 22.5
176/4=44
65/4=16.25
A. 22.5 CM tall, 44 CM wide and 65 CM deep

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5(n+1) equals 2 ( n+ 7)

alina1380 [7]

The wording is confusing but I am going to guess it is 5(n+1)=2(n+7). So 5xn=5n and 5x1=5. 5n+5=2(n+7). Then 2xn=2n and 2x7=14. The question now stands 5n+5=2n+14. Then 5n+2n=7n and 14-5=9. So, you have 7n=9. Then you divide both sides by 7 and get n=1.2857...

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Refrection across the y-axis changes the function to y = |-x|.
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3 years ago

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3 years ago

What would y=x^2 +x+ 2 be in vertex form

balu736 [363]

Answer:

$y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

Step-by-step explanation:

$y = {x}^{2} + x + 2$

We can covert the standard form into the vertex form by either using the formula, completing the square or with calculus.

$y = a(x - h)^{2} + k$

The following equation above is the vertex form of Quadratic Function.

<u>Vertex</u><u> </u><u>—</u><u> </u><u>Formula</u>

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We substitute the value of these terms from the standard form.

$y = a {x}^{2} + bx + c$

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Our h is - 1/2

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<u>Vertex</u><u> </u><u>—</u><u> </u><u>Calculus</u>

We can use differential or derivative to find the vertex as well.

$f(x) = a {x}^{n}$

Therefore our derivative of f(x) —

$f'(x) = n \times a {x}^{n - 1}$

From the standard form of the given equation.

$y = {x}^{2} + x + 2$

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$f'(x) = (2 \times 1 {x}^{2 - 1} ) + (1 \times {x}^{1 - 1} ) + 0$

Any constants that are differentiated will automatically become 0.

$f'(x) = 2 {x}+ 1$

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From the vertex form, our vertex is at (h,k)

Therefore, substitute h = -1/2 and k = 7/4 in the equation.

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2 years ago

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labwork [276]

Answer:

Step-by-step explanation:

equation

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Read 2 more answers