If a farmer ships 4 times as many oranges as tangerines. The farmer ships 8260 oranges. How many tangerines does he ship.

What is -1 and 3/5 divided by -2/3? As a fraction.

Dennis_Churaev [7]

Answer=

mixed form=2 2/5

decimal form=2.4

exact form= 12/5

8 0

3 years ago

Solve each system of equations by graphing. y = 2x y = x + 1

yawa3891 [41]

Step-by-step explanation:

The system of linear equations are , (i) y = 2x and (ii) y = x + 1 . We will find some coordinates and then we will plot its graph. The point where both Graphs will meet will be the solution of the graph .

• Finding coordinates of Equⁿ (i) :-

Step 1 : Put x = 0 :-

$\tt y = 2x = 2(0) = \red{0}$

Step 2: Put x = 1 :-

$\tt y = 2x = 2(1) = \red{2}$

$\boxed{\begin{array}{c|c|c} \bf x & \tt 0 &\tt 1 \\ \bf y &\tt 0 & \tt 2 \end{array}}$

_________________________________

• Finding coordinates of equⁿ (ii) :-

Step 1 : Put x = 0 :-

$\tt y = x + 1 = 1 + 0 = \red{1}$

Step 2: Put x = 1 :-

$\tt y = x + 1 = 1 + 1 = \red{2}$

$\boxed{\begin{array}{c|c|c} \bf x & \tt 0 &\tt 1 \\ \bf y &\tt 1 & \tt 2\end{array}}$

Now plot these points on the graph Taking appropriate scale . Refer to the attachment in graph . From graph the Solution is (1,2) .

8 0

3 years ago

Que How does your tessellation with triangles relate to hexagons and tessellating the plane? Type your response in the space bel

madreJ [45]

For a regular tessellation, the shapes can be duplicated infinitely to fill a plane such that there is no gap. The only shapes that can form regular tessellations are equilateral traingle(all sides are equal. This means that it can be turned to any side and it would remain the same), square and regular hexagon. Looking at the given options, we have

Shape Tessellate?

Octagon No

Hexagon Yes

Pentagon No

Square Yes

Triangle No(unless it is specified that it is an equilateral triangle)

5 0

1 year ago

The city of Orlando’s building a dog park in downtown Orlando dog park will be triangular shaped as shown on the coordinate plan

kiruha [24]

Answer:

Perimeter of the dog park = 15.4 yards

Step-by-step explanation:

Coordinates of the vertices of the given triangle are,

P(1, 2), Q(1, 6), R(-4, 2)

Since distance between the two points $(x_1, y_1)$ and $(x_2,y_2)$ is,

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

Length of PQ = $\sqrt{(1-1)^2+(2-6)^2}$

PQ = 4

Length of PR = $\sqrt{(1+4)^2+(2-2)^2}$

PR = 5

Length of QR = $\sqrt{(1+4)^2+(6-2)^2}$

QR = $\sqrt{25+16}$

= $\sqrt{41}$

= 6.4 yards

Therefore, perimeter of the given triangle = PQ + QR + PR

= 4 + 6.4 + 5

= 15.4 yards

4 0

4 years ago

AleksAgata [21]

$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\textit{sphere A}}{\textit{sphere B}}\qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}\qquad \qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}=\stackrel{\stackrel{volumes'}{ratio}}{\cfrac{\sqrt[3]{288}}{\sqrt[3]{v}}}\implies \cfrac{1}{4}=\sqrt[3]{\cfrac{288}{v}}\implies \left( \cfrac{1}{4} \right)^3=\cfrac{288}{v} \\\\\\ \cfrac{1^3}{4^3}=\cfrac{288}{v}\implies \cfrac{1}{64}=\cfrac{288}{v}\implies v=18432$

3 0

4 years ago