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3 years ago

How do you work it out?

Mathematics

1 answer:

expeople1 [14]3 years ago

8 0

You would first find 25% of 90, which would be 22.5. Then subtract that from 90 minutes, you would then get 67.5. Therefore, it would be about 67 minutes would be used toward the recipe.

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Whats the answer all of them explain how to find which is greather 1/4 of 12 or 1/3 of 12 explain your answer

katrin [286]

Answer:

1/3 of 12 would be greater than 1/4 of 12.

Step-by-step explanation:

One third (1/3) of twelve results in 4. (Multiply 12/1 by 1/3 and simplify.)

One fourth (1/4) of twelve results in 3. (Multiply 12/1 by 1/4 and simplify.)

4 > 3.

Therefore, 1/3 of 12 is greater then 1/4 of 12.

5 0

3 years ago

Find the missing side of each triangle. Leave your answers in simplest radical form.

rosijanka [135]

Answer:

Step-by-step explanation:

pythagoras theorem

a^2+b^2=c^2

8^2+9^2=x^2

64+81=x^2

145=x^2

$\sqrt{145$ =x

4 0

2 years ago

Read 2 more answers

Help me solve this please please

RUDIKE [14]

Answer:

gdrfhuguyf 5w5487657

Step-by-step explanation:

canscZsmzicawlzciwjsdnpoopecuihaisuhdciuhiwhecyoumlsndsuihaweiuc

4 0

2 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

Represent the following expressions as the power of a number a.<br> 1/a^2 · 1/a^3

wolverine [178]

Answer:

a⁻⁵

Step-by-step explanation:

1/a²·1/a³

Multiply numerators and denominators

= (1 × 1)/(a² × a³)

= 1/a⁵

= a⁻⁵

4 0

2 years ago