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Valentin [98]
3 years ago
14

Draw a model to show a fraction that is equivalent to one third

Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0
Hope this helps :) (the first circle is a fraction of thirds and the second circle is a fraction of sixths)

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The simple interest formula is I = Prt, where I represents simple interest on an amount, P, for t years at a rate of r, where r
andrey2020 [161]

Answer:

$80

Step-by-step explanation:

From the question, we are given The simple interest formula as I = Prt,

where l = simple interest on an amount= $40

t= years=5years

at a rate r= 10%

The rate= 10/100= 0.1

Then since we are looking for P, we can make P subject of formula from this expression;

I = Prt

P=I/rt

P= 40/(0.1 ×5)

P= 40/0.5

P= 80

therefore, the amount of money p that is needed is $80

5 0
3 years ago
4x + 1/3 y 2 what is the value of the expression above when x = 2 and y = 3
GenaCL600 [577]
Hoi!

To solve this, first plug in the values for x and y.

x = 2, so anywhere you see x, put 2 in its place.

y = 3, so anywhere you see y, put 3 in its place.

4(2) +  \frac{1}{2}(3)

4 × 2 = 8

\frac{1}{2} × 3 = 1 \frac{1}{2}


8 +1 \frac{1}{2} = 9 \frac{1}{2}

9 \frac{1}{2} is your answer.



4 0
3 years ago
Read 2 more answers
Cynthia is constructing a circle inscribed in a triangle. She has partially completed the construction as shown below. What shou
il63 [147K]
I think the answer is A
5 0
3 years ago
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For which intervals the graphs of the functions f(x) = x^3 + x^2 - 4x - 4 is positive
Nookie1986 [14]

Step-by-step explanation:

Consider a function

f

(

x

)

which is twice differentiable. The graph of such a function will be concave upwards in the intervals where the second derivative is positive and the graph will be concave downwards in the intervals where the second derivative is negative. To find these intervals we need to find the inflection points i.e. the x-values where the second derivative is 0.

5 0
3 years ago
Factor completely 81x8 − 1. (9x4 − 1)(9x4 1) (3x2 − 1)(3x2 1)(9x4 − 1) (3x2 − 1)(3x2 1)(9x4 1) (3x2 − 1)(3x2 1)(3x2 1)(3x2 1).
mojhsa [17]

Answer:

(3x^2 - 1)(3x^2 + 1)(9x^4 + 1).

Step-by-step explanation:

Using the identity for the difference of 2 squares;

a^2 - b^2 = (a - b)(a + b)  

we put a^2 = 81x^8 and b^2 = 1  giving

a = 9x^4 and b = 1, so:

81x^8 − 1 =   (9x^4 - 1)(9x^4 + 1)

Applying the difference of 2 squares to 9x^4 - 1:

= (3x^2 - 1)(3x^2 + 1)(9x^4 + 1).

4 0
2 years ago
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