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sladkih [1.3K]
3 years ago
5

How long is a ruler?​

Mathematics
2 answers:
Dennis_Churaev [7]3 years ago
6 0

Answer:

Depending on what ruler u have a standard one is 12 inches and 30 centimeters.

Gelneren [198K]3 years ago
4 0

Answer:

A ruler is 12 inches (or one foot.)

Step-by-step explanation:

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Dmitriy789 [7]
The answer is 22.222222%.
5 0
4 years ago
Read 2 more answers
Las+salchichas+vienen+en+paquetes+de+6+y+los+panes+para+panchos+se+envasan+en+paquetes+de+8+unidades+¿Es+cierto+que+si+se+prepar
vovikov84 [41]

Answer:

Es cierto se usan 16 paquetes de salchichas y 12 paquetes de panes para pancho.

Step-by-step explanation:

Lo que debemos hacer en este caso es dividir la cantidad total de panchos primero por la cantidad de unidades de salchichas que vienen en su paquete y segundo por la cantidad de unidades de panes para panchos que vienen en su paquete, si el resultado nos da un numero entero podemos decir que usa una cantidad completa de paquetes de salchichas y de panes:

96/6 = 16

96/8 = 12

Es decir que para 96 panchos se usan 16 paquetes de salchichas y 12 paquetes de panes para pancho, por lo tanto es cierto.

8 0
4 years ago
HELPP!!! 2 questions please!
OlgaM077 [116]

Answer:

Step-by-step explanation:

I need help with same question pleas.

5 0
3 years ago
Which option shows the following fractions written with a common<br>denominator?<br>8/9 and 3/8​
Mumz [18]

Answer:

72

Step-by-step explanation:

8/9 + 3/8

Multiply the denominators to get the common denominator

9x8 = 72.

4 0
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}=1y=\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
2 years ago
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