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Likurg_2 [28]
3 years ago
12

Can you guys help me.

Mathematics
1 answer:
Alex Ar [27]3 years ago
6 0
The banner would be 10feet 3inches
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No comprendo. I'm doing a test review and I'm screwed​
MAXImum [283]

Answer:

x = 5

Step-by-step explanation:

sin theta = opposite side / hypotenuse

sin 53 = 4/x

Multiply each side by x

x sin 53 = 4/x *x

x sin 53 = 4

Divide each side by sin 53

x sin 53 / sin 53 = 4 / sin 53

x = 4 /sin 53

x = 5.008542633

To the nearest whole number

x = 5

5 0
4 years ago
Read 2 more answers
8.)What is the Measurement of angle 2?<br><br> 9.) What is the measure of angle 7?
EastWind [94]

Answer:

angle 2 = 90° {corresponding angles}

angle 7 + angle 5 = 180° {linear pair}

or, angle 7 + 100° = 180° {angle 5 is corresponding to 100°}

so, angle 7 = 80°

8 0
2 years ago
A sheet of bad paper is cut into 12 equal parts. What percent of the whole is 3 parts?
11Alexandr11 [23.1K]
25%

Divide 12 by 3 and you'll get 4.
That means 1/4 or a quarter of the 12 is 3.
(Convert the fraction to its percentage form)
1/4 or 1 divided by 4 equals 0.25
Multiply 0.25 by 100 and you'll get 25%
3 0
3 years ago
Read 2 more answers
Um help. I have no clue what to do​
Arlecino [84]

I’m not sure but I think it’s sample x

5 0
3 years ago
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 }\\\\&#10;\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ({{ h}},{{ k}})\\&#10;vertices\ ({{ h}}, {{ k}}\pm a)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{{{ a }}^2+{{ b }}^2}&#10;\end{cases}\\\\&#10;-------------------------------\\\\

\bf \begin{cases}&#10;h=0\\&#10;k=0\\&#10;a=4\\&#10;c=5&#10;\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&#10;\\\\\\&#10;c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b&#10;\\\\\\&#10;\sqrt{5^2-4^2}=b\implies \boxed{3=b}&#10;\\\\\\&#10;\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}

7 0
4 years ago
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