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IrinaVladis [17]

3 years ago

11

the cost of building amphitheater including the stage is 225 a square yard the budget provided to build the amphitheater is 6500

0 are they within budget

1 answer:

Ira Lisetskai [31]3 years ago

6 0

Yes it is hello what’s good

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a paper company needs to ship paper to a large shipping business. The paper will be shipped in small boxes and large boxes. The

Pani-rosa [81]

Answer:
x = 2 y
190 = 10 y + 18 x :
where, x = Number of large boxes, y = number of small boxes
Step-by-step explanation:
Let us assume the number of large boxes shipped = x
So, the number of small boxes shipped = y
⇒ x = 2(times y) = 2 y, or x = 2 y
Now, the volume of 1 small box = 10 cubic feet
So, the volume of y small boxes = y ( Volume of 1 small box)
= y ( 10) = 10 y
Now, the volume of 1 large box = 18 cubic feet
So, the volume of x large boxes = x ( Volume of 1 large box)
= x( 18) = 18 x
Total Volume Transferred = Volume of {Small boxes + Large boxes}
or, 190 cubic feet = 10 y + 18 x
So, the needed system of equations is:
x = 2 y
190 = 10 y + 18 x :
where, x = Number of large boxes, y = number of small boxes

Read more on Brainly.com - brainly.com/question/13862571#readmore

5 0

3 years ago

18) In a class of 50 boys 3/5of the boys got marks above 60 in Mathematics. How

bonufazy [111]

Answer:

thx for the points mwah

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

what is the area of a regular pentagon with sides of length 13 and an apothem of length 8.95, in square units

Assoli18 [71]

The area is 290.760680699535.

3 0

3 years ago

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

4 years ago

PLEASE HELP. MATH IN PICTURE

marin [14]

9514 1404 393

Answer:

D . . . (best of the erroneous choices)

Step-by-step explanation:

Solving the first equation for x, we get ...

√(y -1) ≥ x

Solving the second equation for x, we get ...

x > 3

Substituting for x, we have ...

√(y -1) > 3

y -1 > 9

y > 10

Ordered pairs that are in the solution set will have coordinates ...

x > 3, y > 10

In interval notation that looks like ...

x ∈ (-∞, 3) and y ∈ (10, ∞)

The closest answer choice is the last one.

_____

You will note that x must be strictly greater than 3, so y cannot be equal to 10. The offered choice is in error on that point.

__

You will also note that y is dependent on x. That is, one cannot pick a value of y greater than 10 independently of the value of x. In that sense, the solution is not "the set of all ordered pairs such that [x and y have independent limits]". Rather, it is the set of all ordered pairs such that √(y -1) ≥ x > 3.

5 0

3 years ago