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3 years ago

Which equation is equivalent to y-34=x(x-12)

Mathematics

1 answer:

egoroff_w [7]3 years ago

7 0

The final answer is:
y - 34 = x2 - 12x

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A trapezoid has an area of 50 square units. The lengths of the bases are 10 units and 15 units. find the height.

Lera25 [3.4K]

Step-by-step explanation:

area = ( base 1 + base 2 ) ÷ 2 x height

50 = ( 10 + 15) ÷2 x height

50 ÷ 12.5 = height

height = 4

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3 years ago

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PolarNik [594]

Answer:

SA = 384 in²

Step-by-step explanation:

Assuming we're finding surface area:

SA of top of large cube: 8(8) - 4(4) = 48 in²

SA of sides of large cube: 4(8)(8) = 256 in²

SA of exposed faces of small cube: 5(4)(4) = 80 in²

Total = 48 + 256 + 80 = 384 = ∑SA

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2 years ago

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3 years ago

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3 years ago

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

1 year ago