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igor_vitrenko [27]
2 years ago
11

Maribella buys 5 pounds of white potatoes and as many pounds of sweet potatoes. How many pounds of sweet potatoes does she buy?

Use paper to
model the problem with a tape diagram. In the boxes below, express your answer as a both a fraction greater than one and a mixed number.
Fraction Greater than One: Maribella buys
pounds of sweet potatoes.
Mixed Number: Maribella buys
pounds of sweet potatoes.
Hey
Mathematics
2 answers:
const2013 [10]2 years ago
3 0

Answer:

8 Pounds of potatoes

vazorg [7]2 years ago
3 0
8 pounds of sweet potatoes
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vredina [299]

Step-by-step explanation:

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3 years ago
Triangle FGH has the following side lengths: FG = 5 ft, GH = 10 ft, HF = 12 ft Triangle PQR is similar to triangle FGH. The long
7nadin3 [17]
The first thing we must do for this case is find the scale factor.
 We have then that for the larger side of both triangle, the scale factor is:
 k =  \frac{RP}{HF}
 k = \frac{7.2}{12}
 k = 0.6
 To find the other two sides, we must apply the scale factor on each side of the triangle FGH.
 We have then:

 For PQ
 
PQ = k * FG

PQ = 0.6 * 5

PQ = 3

 For QR
 
QR = k * GH

QR = 0.6 * 10

QR = 6

 Answer:
 
You have that the lengths for the other two sides of triangle PQR are:
 
PQ = 3

QR = 6
3 0
3 years ago
What is the best name to describe an object that has four right angles, the opposite sides are parallel, and it has four equal s
MAXImum [283]

Answer:

D-Square

Step-by-step explanation:

That describes a square, so the answer a square.

6 0
3 years ago
Read 2 more answers
Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter
Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and  \sqrt{8^{2} - x^{2} }or  \sqrt{64 - x^{2} }

thus tan(arcsin(x/8))  = \frac{x}{\sqrt{64 - x^{2} }}  =

Therefore  ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B

Here we have cos = adjacent/hypotenuse where adjacent side is \sqrt{64 - x^{2} } and hypothenuse = 8 we have \sqrt{64 - x^{2} }/8

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

4 0
3 years ago
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