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3 years ago

Answer plzzz. Tell the truth and answer correctly

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

3 0

Answer:

54 feet

Step-by-step explanation:

We can use proportions

1 inch 27 inches

--------- = ------------

2 feet x feet

Using cross products

1x = 54

The building is 54 feet

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Tim install 50 square feet of his floor in 45 mins. At this rate how long does it take him to install 495 square feet. Is that 9

Nimfa-mama [501]

Answer:

the answer is 445.5

Step-by-step explanation:

what I did was divide 50 by 495 and got 9.9 then since it took 45 minutes every 50 square feet of floor i multiplied 45 x 9.9 and got 445.5

6 0

3 years ago

Work out the sum of

Sladkaya [172]

Answer: 1 and 3/70

Explanation: Find the least common denominator and then combine like terms. The LCM is 70.

8 0

3 years ago

Ava is 23 cm taller than Olivia, and Olivia is half the height of Lucas. If Lucas is 1.78 m tall, how tall are Ava and Olivia? E

Colt1911 [192]

Ava is 1.09 and Olivia is 0.86

3 0

3 years ago

Both Need To Be Checked :-)

Tems11 [23]

I think u should check the first one again u made a mistake but the second one is correct if u need any more help ask me

6 0

2 years ago

Read 2 more answers

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

3 years ago