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2 years ago

Can someone solve this? The answer needs to be in fraction form so can't be done by a calculator.

Mathematics

1 answer:

Lesechka [4]2 years ago

7 0

Answer:

9 1/6

You may use a fraction calulator for this question by the way!

Hope this helps!

Sorry if its wrong!

Have a nice day!

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Seraphina says that angle KLM is a right triangle. Is she correct? Explain.

katovenus [111]

Answer:

Incorrect

Step-by-step explanation:

The angle is acute,which means under right triangle's usual size.

You need at least 90 cm to have a right triangle so no she is incorrect

4 0

2 years ago

Read 2 more answers

3) Which ordered pair in the form (x, y) is a solution of this equation?<br> 7x - 5y = 1

Ludmilka [50]

(x,y) = (1, 1.2)
1 = 7(1) - 5(1.2)
1 = 7 - 6

There are many other options though.

3 0

2 years ago

Molly was curious if quadrilaterals ABCDABCDA, B, C, D and EFGHEFGHE, F, G, H were congruent, so she tried to map one figure ont

Novay_Z [31]

Answer:

The answer is below

Step-by-step explanation:

A transformation of a point is the movement of an point from an initial position to a new position. If an object is transformed, all the point of an object is transformed. Two figures are said to be congruent if they have the same shape and the measure of each of their sides are the same.

An object is congruent to another object if the object can be mapped to the other object when transformed.

6 0

2 years ago

How much will the taxi driver earn if he takes one passenger 4.8 miles and another passenger 7.3 miles? Explain your process.

Bezzdna [24]

Answer:

35.225 I believe

Step-by-step explanation:

You can always check on Slader.

4 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid

Artist 52 [7]

Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.

$\bf \textit{volume of a rectangular prism}\\\\
V=lwh\quad 
\begin{cases}
l = length\\
w=width\\
h=height\\
-----\\
w=l=x
\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\
-------------------------------\\\\
\textit{surface area}\\\\
S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h
\\\\\\
\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\
-------------------------------\\\\
V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3$

so.. that'd be the V(x) for such box, now, where is the maximum point at?

$\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2
\\\\\\
\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100
\\\\\\
x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}$

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.

so, the volume will then be at $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$

6 0

2 years ago