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Alex_Xolod [135]

4 years ago

13

What can you do to make 50 into 15. And do the same thing to 60 so they can both be divided equally because they are not the sam

e number at the end.

1 answer:

Virty [35]4 years ago

4 0

Simplify the numbers
like...
5 10 15 20 25 30 35 40 45 50 55 60

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A batch of snack mix made of dried fruit and nuts contain 744G of nuts the ratio of nuts to dried fruit by weight is 12 to 13 ho

mixer [17]

In a bag of snack mix:

n = nuts d = dried fruit

n = 744g d = ???g

Ratio Nuts/Dried fruit = 12/13

This basically means that if the mix was divided in 25 parts, 12 parts would be nuts and 13 would be dried fruits

If 744 is 12 parts then 744/12 is 1 part

744/12 = 62

1 part = 62g

there are 13 parts of dried fruit

13 x 62 = 806

744g of nuts + 806g of dreid fruit = 1550g

Answer: A batch of snack mix weigh 1550g, or 1.55kg

hope it helps :)

5 0

4 years ago

Determine the area of the rhombus base 6 height 4

Leno4ka [110]

48
There are two sets of equal triangles together they each equal a square so one sets area is 36 and the others area is 12

6 0

3 years ago

You can always _____ two similar triangles so that they become congruent.

storchak [24]

Answer:

<em>t</em><em>h</em><em>e</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>i</em><em>s</em><em> </em><em>h</em><em>a</em><em>v</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>b</em><em>e</em><em> </em><em>e</em><em>q</em><em>u</em><em>a</em><em>l</em><em>.</em>

Step-by-step explanation:

you can always equal two similar triangles so that they become congruent.

In the ratio of the length of their sides is equal to one, they are congruent. So, all congruent triangles have to be similar but all similar triangles need not be congruent......the ratio of corresponding sides,AB/A'B'=BC/B'C'=CA/C'A'. If the two triangles are congruent AB/A'B'=BC/B'C'=CA/C'A'=1

I am glad to help you!!!!

5 0

3 years ago

Find lim h->0 f(9+h)-f(9)/h if f(x)=x^4 a. 23 b. -2916 c. 2916 d. 2925

Svetach [21]

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h$

Carry out the binomial expansion in the numerator:

$(9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4$

Then the 9⁴ terms cancel each other, so in the limit we have

$\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h$

Since <em>h</em> is approaching 0, that means <em>h</em> ≠ 0, so we can cancel the common factor of <em>h</em> in both numerator and denominator:

$\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)$

Then when <em>h</em> converges to 0, each remaining term containing <em>h</em> goes to 0, leaving you with

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}$

or choice C.

Alternatively, you can recognize the given limit as the derivative of <em>f(x)</em> at <em>x</em> = 9:

$f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h$

We have <em>f(x)</em> = <em>x</em> ⁴, so <em>f '(x)</em> = 4<em>x</em> ³, and evaluating this at <em>x</em> = 9 gives the same result, 2916.

8 0

3 years ago

How to use the binomial therom for (2x + 5)^5

Nady [450]

Answer:

32

x

5

−

400

x

4

+

2000

x

3

−

5000

x

2

+

6250

x

−

3125

Step-by-step explanation:

8 0

3 years ago