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3 years ago

Can someone plz help me thxxxx

Mathematics

1 answer:

maria [59]3 years ago

6 0

Yes, rectangular prisms may have different heights but the same volume.

For instance a rectangular prism may be : 7×5×5
Another way to have this same product and have different heights is : 7×25

Hope this helped - Maria S.

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7(C - 18)= -98<br> Help ASAP

Harlamova29_29 [7]

Answer:

$C=4$

Step-by-step explanation:

$7(c-18)=-98\\7c \frac{-126}{+126}=\frac{-98}{+126} \\\frac{7c}{7}=\frac{28}{7}\\c=4$

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3 years ago

A pet store can put no more then 5 fish in each tank. What is the least number of tanks the store needs to hold 354 fish?

Vlada [557]

70.8 fish tanks. Because you have to divide 354 by 5.

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3 years ago

WITH AN EQUATION

andrey2020 [161]

Answer:

x-35.75=40.00

Step-by-step explanation:

Let's use <em>m </em> as our variable.

So Marlen went to the mall and spent $35.75 on shoes. When she got home she had $40.00 in her purse.

An equation to represent this is : x-35.75=40.00

If you are interested in the value of x it is $75.75

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3 years ago

A pole that is 3.4 m tall casts a shadow that is 1.54 m long. At the same time, a nearby building casts a shadow that is 35.75 m

inna [77]

Answer:

The building would be 78.92 meters tall.

Step-by-step explanation:

If the shodow of a 3.4m tall pole is 1.54m tall, and you have the shadow's height, then you make a cross, multiply-divide ratio table.

5 0

3 years ago

What sine function represents an amplitude of 1, a period of 2π, a horizontal shift of π, and a vertical shift of −4?

Gwar [14]

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

$\bf % template detailing
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

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

with that template in mind, let's see

the period of 2π, well, for that, you have to do nothing, because that's sine original period

Amplitude of 1, well, that's also sine's original amplitude

horizontal shift/phase... ok, that means C/B = π, now, it doesn't say if to the left or the right, so I gather either is ok, so...let's do the right

you could use then, C = -π and B = 1, that gives you -π/1 or -π

vertical shift of -4, well, that simply means D = -4

f(θ) = sin(θ - π) - 4 <-- will do

5 0

4 years ago