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Nana76 [90]
3 years ago
7

What is the scale factor from ABC to DEF? A. 5 B. 1 C. 0 D. 10​

Mathematics
1 answer:
natali 33 [55]3 years ago
8 0

Answer:

b

Step-by-step explanation:

they both have 5 so we divide 5/5 and we get one.

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Write the inequality:<br> 3 less than n is no more than -8
rusak2 [61]

Answer:

n - 3 ≤ -8

Step-by-step explanation:

n - 3 ≤ -8

3 less than n:  n - 3

is no more than: ≤ - 8

5 0
2 years ago
Use the Distributive Property to multiply.) -4x(x - 7)*
ad-work [718]

Answer:

-4x^2-28x

Step-by-step explanation:

Distribute

-4x(x) and -4x(-7)

-4x^2 and -28x

5 0
3 years ago
Read 2 more answers
5(1-cos x)÷x<br> Determine the limit of the transcendental function (if it exists).
luda_lava [24]

hello here is a solution :

8 0
3 years ago
X squared plus 3X -4 equals six
professor190 [17]
X^2 + 3x - 4 = 6
x^2 + 3x - 10 = 0 (Now reverse FOIL)
(x + 5)(x - 2)(Now x will equal the opposites of your numbers)
x = -5
x = 2
Hope this helped!
5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0
3 years ago
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