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3 years ago

Five cats each ate 1/4 cup of cat food. Four other cats each ate 1/3 cup of cat food how much food did the nine cats eat?

Mathematics

2 answers:

makkiz [27]3 years ago

6 0

They all ate 7/12 cup of catfood. Add the numbers to find the total :>

suter [353]3 years ago

6 0

Answer:

2 and a half cups of food

Step-by-step explanation:

You might be interested in

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

Describe the translation of the function from the parent function f(x) = x².

kkurt [141]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\

\end{array}\\$

$\bf \begin{array}{llll}
\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}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}
$

now, notice the template above... now let's see your function $\bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}$

so.. what do you think was the shift then?

8 0

3 years ago

PLEASE HELP Solve |10x + 15| >= 105

yan [13]

$|10x + 15| \geq 105\\
 5|2x + 3| \geq 105\\
|2x+3|\geq21\\
2x+3\geq21 \vee 2x+3\leq-21\\
2x\geq 18\vee2x\leq-24\\
x\geq9 \vee x\leq-12\\
x\in(-\infty,-12\rangle\cup\langle9,\infty)$

8 0

3 years ago

A certain company's main source of income is a mobile app. The company's annual profit (in millions of dollars) as a function of

liberstina [14]

Answer:

x = $3, or x = $11

Step-by-step explanation:

The equation given is $P(x)=-2(x-3)(x-11)$

where

P(x) is the profit, and
x is the app price

We want app prices (x's) when profit (P(x)) is 0, so plugging in into the equation:

$P(x)=-2(x-3)(x-11)\\0=-2(x-3)(x-11)\\0=(x-3)(x-11)$

It means (x-3) = 0 OR (x-11) = 0

So, x = 3, or 11

7 0

3 years ago

Read 2 more answers

You and some friends are going to the fair. Small rides costs 4 tickets and while big rides costs 7 tickets. You and your friend

tensa zangetsu [6.8K]

Answer:

4x+7y≤250

Step-by-step explanation:

Inequalities

Let's call:

x = number of small rides

y = number of big rides

Since each small ride costs 4 tickets, x rides cost 4x tickets.

Since each big ride costs 7 tickets, y rides cost 7y tickets.

The total number of tickets is the sum of both:

Total tickets = 4x+7y

The friends have 250 tickets, so the total tickets of the rides cannot be greater than the 250 tickets available, thus:

4x+7y≤250

7 0

3 years ago