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Ksenya-84 [330]
3 years ago
5

⚠️Tell if the following systems of equations have zero, one, or infinite solutions: -6x + 3y=18 and y=2x - 2⚠️

Mathematics
1 answer:
fgiga [73]3 years ago
4 0

Answer:

0 solutions

Step-by-step explanation:

-6x+3y=18

  y=2x-2.   so substitute y in the first equation

-6x+3(2x-2)=18 distribute and combine like terms

-6x +6x-6=18 and we end up with

-6=18 witch is never true so the system of equation has 0 solutions

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Find the missing side length. Round to the nearest tenth.<br>15 cm<br>9 cm​
vazorg [7]
Using Pythagorean’s theorem
a^2 +b^2= c^2
We know
c=15
b=9
a=?
Sub that in the Pythagorean’s theorem
a^2+9^2 = 15^2
a^2 + 81 = 225
a^2= 225-81
a^2 = 144
a= square root of 144
a= 12

Therefore the answer is 12
5 0
3 years ago
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Papessa [141]

Answer:

Step-by-step explanation:

1)First convert mixed fraction to improper fraction and them prime factorize

6\frac{1}{4} = \frac{25}{4}\\

\sqrt{\frac{25}{4}}= \sqrt{\frac{5*5}{2*2}}= \frac{5}{2} = 2 \frac{1}{2} \\\\

2)

(2 \frac{1}{2}- 1 \frac{1}{2})*1 \frac{1}{7}=( \frac{5}{2}- \frac{3}{2})* \frac{8}{7}\\\\\\= \frac{2}{2}* \frac{8}{7}\\\\=1* \frac{8}{7}= \frac{8}{7}\\\\\\=1 \frac{1}{7}

3) 0.00706 = 7.06 * 10^{-3}

<em>4)</em> 144 = 12 * 12

12 = 6*2

6 = 2*3

Prime factorization of 144 = 2 * 3 * 2 * 2 * 3 *2

                                          = 2⁴ * 3²

5) To find LCM, prime factorize 96 & 144

96 = 2 * 2 * 2 * 2 * 2 * 3 = 2⁵ * 3

144 = 2⁴ * 3²

LCM = 2⁵ * 3² = 32 * 9 = 288

6) HCF

105 = 7 * 5 * 3

135 = 5 * 3* 3 * 3

180 = 5 * 3 * 3 * 2 * 2

HCF = 5 * 3 = 15

7) 24 = 3 * 2 * 2 * 2 = 3 * 2³

   36 = 3 * 3 * 2 * 2 = 3² * 2²

   40 = 5 * 2 * 2 * 2 = 5 * 2³

LCM = 5 * 2³ * 3² = 5 * 8 * 9  = 360

HCF = 2² = 4

Difference = 360 - 4 = 356

8) Multiply each digit of the binary number by the corresponding power of 2, solve the powers and add them all

1111 = 1 *2³ + 1*2² + 1*2¹ + 1*2° = 8 + 4 + 2 + 1 = 15

Ans: 15

9) 36₇ = 102₅

10) 6.9163 = 6.916

6 0
4 years ago
Read 2 more answers
HELP ME PLEASE!
Delicious77 [7]
Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides).  If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.

Let's see whether they do.

8^2 + 8^2 = 11^2  ???
 64    + 64 = 121?  NO.  This is not a right triangle.

If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:

8^2 + 8^2 = d^2, where d=hypotenuse.  Then 64+64 = d^2, and 

d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.

11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.
7 0
3 years ago
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Rama09 [41]

Answer:

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Step-by-step explanation:

34 = -62 + 3x

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5 0
4 years ago
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Andreyy89

Answer:

B

Step-by-step explanation:

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6 0
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