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2 years ago

The cost of 6sandwiches and 4 drinks is $53. The cost of 4 sandwiches and 6 drinks is $47. How much does one sandwich cost?

Mathematics

1 answer:

vazorg [7]2 years ago

5 0

Answer:

$8.00

Step-by-step explanation:

6 divided by 53=8

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Find the length of the diagonal, round to the nearest tenth (( please answer, ill give brainliest))

Lilit [14]

Answer:

Length of diagonal in the first top figure is 13 yd

Okay for this one, the formula of a diagonal in a cuboid is the root of sum of squares

length, breadth and height

D = root of 12^2 + 4^2 + 3^2

D = root of 144 + 16 + 9

D = root 169

D = 13yd

Length of diagonal in the bottom figure is 10.77 which is 10.8m

For this one, you have to find the diameter of base first

Since radius of the base is 5 diameter = 10

Since its a right angled triangle, Hypotenuse square = sum of sides square

Diagonal^2 = 10^2 + 4^2

D^2 = 100 + 16

D^2 = 116

D = root 116

D = 10.77

D = 10.8 m

6 0

3 years ago

For every batch of cookies Carla's made , her brother ate 2 cookies. If Carla's brother ate a total of 6 cookies, how many batch

spayn [35]

Step-by-step explanation:

number of the total cookies ÷ the nber her brother ate

6÷2=3

3 0

3 years ago

How many terms are there in a geometric series if the first term is 4, the common ration is 3 and the sum of the series is 160

Marina CMI [18]

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=4\\
r=3\\
S_n=160
\end{cases}$

$\bf 160=4\left( \cfrac{1-3^n}{1-3} \right)\implies 160=4\left( \cfrac{1-3^n}{-2} \right)\implies 160=-2(1-3^n)
\\\\\\
160=2(3^n-1)\implies \cfrac{160}{2}=3^n-1\implies 80=3^n-1
\\\\\\
81=3^n~~
\begin{cases}
81=3\cdot 3\cdot 3\cdot 3\\
\qquad 3^4
\end{cases}\implies 3^4=3^n\implies 4=n$

3 0

3 years ago

Square OABC is drawn on a centimetre grid. O is (0,0) A is (3,0) B is (3,3) C is (0,3) Write down how many invariant points ther

storchak [24]

Answer:

No invariant point

Step-by-step explanation:

Hello!

When we translate a form, in this case a polygon We must observe the direction of the vector. Since our vector is:

$\vec{v}=\left ( \begin{matrix}1\\ 3\end{matrix} \right )$

1) Let's apply that translation to this polygon, a square. Check it below:

2) The invariant points are the points that didn't change after the transformation, simply put the points that haven't changed.

Examining the graph, we can see that no, there is not an invariant point, after the translation. There is no common point that belongs to OABC and O'A'B'C' simultaneously. All points moved.

3 0

3 years ago

The points A(–5, 5) and B(–5, –7) are plotted on the coordinate plane.

Ghella [55]

C I did this before

7 0

3 years ago