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2 years ago

(5x - 7) + (10x 7) = 180

Mathematics

1 answer:

goldfiish [28.3K]2 years ago

4 0

15x=180
x=12

hope this helps

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Tyler went to the supermarket to buy food for a food pantry. He has $36, and can carry up to 20 pounds of food in his backpack.

sattari [20]

Answer:

Solutions: (2,10), (4,5)

Not solutions: (1,12), (6,10), (12,8), (18,6)

Step-by-step explanation:

Let x be the number of packages of pasta and y be the number of jars of pasta sauce. If pasta costs $1 for a 1-pound package, then x packages of pasta cost $x and weigh x pounds. If pasta sauce costs $3 for a 1.5 pound jar, then y jars cost $3y and weigh 1.5y pounds.

1. Tyler has $36, then

$x+3y\le 36.$

2. Tyler can carry up to 20 pounds of food in his backpack, then

$x+1.5y\le 20.$

You get the following system of inequalities:

$\left\{\begin{array}{l}x+3y\le 36\\ x+1.5y\le 20\end{array}\right.$

Now substitute the coordinates of each point:

(1,12):

$\left\{\begin{array}{l}1+3\cdot 12=37> 36\\ 1+1.5\cdot 12=19\le 20\end{array}\right.$

False, because first inequality doesn't hold.

(2,10):

$\left\{\begin{array}{l}2+3\cdot 10=32\le 36\\ 2+1.5\cdot 10=17\le 20\end{array}\right.$

True, both inequalities hold.

(4,5):

$\left\{\begin{array}{l}4+3\cdot 5=19\le 36\\ 4+1.5\cdot 5=11.5\le 20\end{array}\right.$

True, both inequalities hold.

(6,10):

$\left\{\begin{array}{l}6+3\cdot 10=36\le 36\\ 6+1.5\cdot 10=21> 20\end{array}\right.$

False, because secondt inequality doesn't hold.

(12,8):

$\left\{\begin{array}{l}12+3\cdot 8=36\le 36\\ 12+1.5\cdot 8=24> 20\end{array}\right.$

False, because second inequality doesn't hold.

(18,6):

$\left\{\begin{array}{l}18+3\cdot 6=36\le 36\\ 18+1.5\cdot 6=27> 20\end{array}\right.$

False, because second inequality doesn't hold.

7 0

2 years ago

Number 1d please help me analytical geometry

lesantik [10]

For a) is just the distance formula

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
% (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}$
-----------------------------------------------------------------------------------------
for b) is also the distance formula, just different coordinates and distance

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
% (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}$
--------------------------------------------------------------------------
for c) well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is $\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
% (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}$

now.. whatever that is, is = BC, so the distance for BC is

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
% (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}$

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d) we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
% (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\
$

$\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}$

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"

7 0

3 years ago

Can someone please help me with a few math questions regarding slope, slope intercept, linear equations etc etc. I'll give 50 po

ELEN [110]

$3.50 when you step into the cab,
and then for however miles you drive add 2.25x

6 0

2 years ago

Read 2 more answers

Ccompares 3.67（）36.7

GalinKa [24]

Answer:

36.7 > 3.67

Step-by-step explanation:

5 0

2 years ago

The ratio of the perimeters of two similar triangles is 4:3. What are the areas of these triangles if the sum of their areas is

emmasim [6.3K]

Answer:

the areas of these triangles are 83.2cm² and 46.8cm²

Step-by-step explanation:

1. If the triangles are similar and the ratio of the perimeter ois 4:3, then the areas are in the following ratio:

4²:3²

16:9

2. The sum of their areas is 65 cm², then, you can calculate the area of the larger triangle as following:

130(16/16+9)

130(0.64)

=83.2cm²

3. The area of the smaller triangle is:

130(9/16+9)

130(0.36)

46.8cm²

Hope this helps

6 0

2 years ago