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3 years ago

What two numbers multiply to 20 and add up to 9

Mathematics

2 answers:

prohojiy [21]3 years ago

3 0

Answer:

5 and 4

Step-by-step explanation:

vova2212 [387]3 years ago

3 0

Answer:

4 and 5

Step-by-step explanation:

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Relationships… are they proportional? !!!20 POINTS!!!

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2 years ago

These figures are congruent what series of transformations moves pentagon ABCDE onto pentagon A’B’C’D’E’?

lora16 [44]

Answer: rotation,translation

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2 years ago

Read 2 more answers

A)x 1/9

STALIN [3.7K]

Answer:

Answer should be x^9

Step-by-step explanation:

This equation looks really complicted, but it's actually much easier when you break it down! First, your going to multiply the fraction 3/2 by 6 - since one is a fraction, youre going to find the GCF, or Greatest Common Factor, and reduce it. The GCF in this equation is 2, so we eliminate the two from the fraction (making it just 3) and divide 6 by 2 (getting 3). Thus, we are left with (x^3)^3 -> 3 x 3 = 9. So we are left with x^9. I hope this helps!

5 0

3 years ago

Select the correct answer. Simplify the expression.<br> (see the screenshot)

Sliva [168]

Answer:

Option D

Step-by-step explanation:

Given expression has been given as,

$\sqrt[5]{224x^{11}y^8}$

$\sqrt[5]{224x^{11}y^8}=\sqrt[5]{2\times 2\times 2\times 2\times 2\times 7(x^{11})(y^8)}$

$=\sqrt[5]{(2^5)\times (7)(x^{10}\times x)(y^5\times y^3)}$

$=2^{\frac{5}{5}}\times 7^{\frac{1}{5}}\times x^{\frac{10}{5}}\times x^{\frac{1}{5}}\times y^{\frac{5}{5} }\times y^{\frac{3}{5} }$

$=2\times 7^{\frac{1}{5}}\times x^2\times y\times x^{\frac{1}{5} }\times y^{\frac{3}{5} }$

$=2x^2y\sqrt[5]{7xy^3}$

Option D will be the answer.

7 0

3 years ago

PLEASE HELP!<br>Calculate $40^{13} \pmod{85}.$

SVEN [57.7K]

In this type of calculations, we decompose 13 by checking the lowest powers of the base, that is 40. for example we check 40^2, or 40^3 and compare it to 85

Notice

40*40*40=64,000

so we check how many time does 85 fit into 64,000:

64,000/85=752.94

85*753=64,005; 64000-64,005=-5

this means that

$40^{3} =-5\pmod{85}$

thus

$40^{13} =40^{3*4+1}={(40^{3})}^{4}*40=(-5)^{4}*40 \pmod{85}=\\\\625*40\pmod{85}=(7*85+30)*40\pmod{85}=30*40\pmod{85}\\\\=1200\pmod{85}=(14*85+10)\pmod{85}=10\pmod{85}$

Answer: 10 (mod85)

Remark, the set of all solutions is:

{......-75, 10, 95, .....}, that is 85k +10

7 0

3 years ago