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pochemuha
2 years ago
7

Can somebody please help me with questions 9 and 10, please

Mathematics
2 answers:
ExtremeBDS [4]2 years ago
6 0
The picture does not show anything, but feel free to message me on what you need help with !
rjkz [21]2 years ago
3 0
We can’t see those questions we don’t know what you are working on.
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1. One liter is equal to 1,000<br> milliliters. How many liters are<br> equal to 18,045 milliliters?
aksik [14]

Answer:

18.045

Step-by-step explanation:

because you will have 18liters and because of 045 you would let it stay the same and instead of a coma you would put a period.

hop this helps

4 0
2 years ago
A hypotenuse of a right triangle is 52 in. One leg of the triangle is 8in. more than twice the length of the other. What is the
DaniilM [7]

c=52, a=2b+8

<span>
Pythagorean Theorem = a^2+b^2=c^2</span>

<span>
(2b+8)^2+b^2=52^2</span>

<span>
<span> 4b^2+32b+64+b^2=2704</span></span>

<span><span>
5b^2+32b-2640=0 </span></span>

<span>
b=20 </span>

a= 2(20)+8 = 48

<span>48+20+52 = 120</span>


3 0
3 years ago
Read 2 more answers
judy makes balloon sculptures at the circus. in 180 minutes, she uses 252 balloons to make 36 identical balloons sculptures, how
xxMikexx [17]
5 minutes to make one ballon sculpture
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3 years ago
Thomas has $1,000 to spend on a vacation. His plane ticket costs $440.
AnnZ [28]
Thomas can spend $118.50 per day
3 0
3 years ago
Which is the equation of a hyperbola with directrices at y = ±2 and foci at (0, 4) and (0, −4)?
Advocard [28]

Answer:

Step-by-step explanation:

First off, I'm assuming that when you said "directrices" you mean the oblique asymptotes, since hyperbolas do not have directrices they have oblique asymptotes.

If we plot the asymptotes and the foci, we see that where the asymptotes cross is at the origin. This means that the center of the hyperbola is (0, 0), which is important to know.

After we plot the foci, we see that they are one the y-axis, which is a vertical axis, which means that the hyperbola opens up and down instead of sideways. Knowing those 2 characteristics, we can determine that the equation we are trying to fill in has the standard form

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1

We know h and k from the center, now we need to find a and b. Those values can be found from the asymptotes. The asymptotes have the standard form

y = ±\frac{a}{b}(x-h)+k

Filling in our asymptotes as they were given to us:

y = ±\frac{2}{1}(x-0)+0 where a is 2 and b is 1. Now we can write the formula for the hyperbola!:

\frac{(y-0)^2}{4}-\frac{(x-0)^2}{1}=1 which of course simplifies to

\frac{y^2}{4}-\frac{x^2}{1}=1

4 0
4 years ago
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