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nignag [31]
3 years ago
12

A rectangle measures 2 1/5 inches by 2 4/5 inches. What is its area?

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
6 0

Answer:

Step-by-step explanation:

20+30?esa es la respuesta

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PLEASE HELP!!
VladimirAG [237]

Answer:

Sphere Volume = (4/3) * PI * radius^3

radius = cube root (Sphere Volume) / (4/3)*PI

radius = cube root (5,600) / (4/3)*PI

radius = 11.016 cm

radius = 11 cm (rounded)

Answer is B

Step-by-step explanation:

4 0
3 years ago
1+1 then divide by 8
Gemiola [76]

Answer:

0.25

Step-by-step explanation:

4 0
2 years ago
There were 3 apple, 1 mixed fruit, 2 grape, and 4 tropical fruit juice boxes in the cooler at the picnic. What is the probabilit
goldenfox [79]

There are  10  juice boxes in the cooler altogether.
2  of them are grape.

The first time Jill pulls one out with her eyes closed,
the probability that it's a grape is

                                         2 / 10 .

If that try is successful, then there are  9  boxes left in the cooler,
and one of them is grape.

If she already has one grape, and reaches in again with her eyes
closed, the probability of pulling out the second grape is

                                        1 / 9 .

The probability of both events happening in two tries is

                           (2/10) x (1/9) = 2/90 = 1/45  =  (2 and 2/9) percent .

6 0
3 years ago
Read 2 more answers
Select the correct answer.
mamaluj [8]

Answer:

C.$0.60

hope this help!!!!!!!!!!!!!!!!!!!!!!

3 0
2 years ago
Carlos drew a plan for his garden on a coordinate plane. Rose bushes are located at A(–5, 4), B(3, 4), and C(3, –5)
BartSMP [9]

Given:

A(-5,4)

B(3,4)

C(3,-5)

So point D is:

so point D is (-5,-5)

For AB is

Distance between two point is:

\begin{gathered} (x_1,y_1)and(x_2,y_2) \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}

so distance between A(-5,4) and B(3,4) is:

\begin{gathered} D=\sqrt[]{(3-(-5))^2+(4-4)^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}

So AB is 8 unit apart.

For B(3,4) and C(3,-5).

\begin{gathered} D=\sqrt[]{(3-3)^2+(-5-4)^2} \\ =\sqrt[]{0^2+(-9)^2} \\ =9 \end{gathered}

So BC is 9 unit apart.

For fourth bush point is (-5,-5) it left of point C(3,-5) is:

\begin{gathered} D=\sqrt[]{(3-(-5))^2+(-5-(-5))^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}

so fourth bush is 8 unit left of C.

For fourth bush(-5,-5) below to point A(-5,4)

\begin{gathered} D=\sqrt[]{(-5-(-5))^2+(4-(-5))^2} \\ =\sqrt[]{0^2+9^2} \\ =9 \end{gathered}

so fourth bush 9 units below of A.

8 0
1 year ago
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