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lys-0071 [83]
4 years ago
15

Which systems of linear equations have no solutions

Mathematics
1 answer:
zhuklara [117]4 years ago
7 0

Answer:

Option A and C

bcoz for no solution.

A1/A2=B1/B2≠C1/C2

where both option is applied in this form

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Apply the distributive property to factor out the greatest common factor of all three terms. {10a - 25 + 5b} =10a−25+5b =
Rom4ik [11]

Answer:

5(2a -5 + b)

Step-by-step explanation:

(10a - 25 + 5b) = 5( 2a - 5 + b)

5(b +  2a  - 5) = 5(2a - 5 + b)

8 0
3 years ago
What is cos 0 when sin 0
taurus [48]

Answer:

Cos θ = √7/3

Step-by-step explanation:

From the question given above, the following data were obtained:

Sine θ = √2 / 3

Cos θ =?

Recall

Sine θ = Opposite / Hypothenus

Sine θ = √2 / 3

Thus,

Opposite = √2

Hypothenus = 3

Next, we shall determine the Adjacent. This can be obtained as follow:

Opposite = √2

Hypothenus = 3

Adjacent =?

Hypo² = Adj² + Opp²

3² = Adj² + (√2)²

9 = Adj² + 2

Collect like terms

9 – 2 = Adj²

7 = Adj²

Take the square root of both side

Adjacent = √7

Finally, we shall determine the value Cos θ. This can be obtained as follow:

Adjacent = √7

Hypothenus = 3

Cos θ =?

Cos θ = Adjacent / Hypothenus

Cos θ = √7/3

4 0
3 years ago
Write the following equations in slope intercept form
Rus_ich [418]

Answer:

problem 1 2x+5y= 0

×=-1/2-5/2y

problem 2 y=1 3/7y y=-1.428571

7y=-10

problem 3 x=-7

problem 4 x-y=20

x=-20

7 0
4 years ago
A liter of water contains about 3.35•10^25 molecules. A certain river discharges about 2.3•10^7 litter of water every second. Ab
dimaraw [331]

Answer:

Jaylynn draws a hen with a scale of

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22 units on her graph paper represents

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6cm6, space, c, m. The hen is

1

6

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A hen on graph paper. A 16-unit long line is labeled, "height." A 2-unit long scale is labeled, "6 centimeters."

6

cm

6 cmheight

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cm

cmc, m, of the actual hen?

Step-by-step explanation:

7 0
3 years ago
Write the first five terms of the the sequence defined by the explicit formula an=72(1/3)^n-1
lisabon 2012 [21]

a_n=72\left(\dfrac{1}{3}\right)^{n-1}\\\\\text{Put}\ n=1,\ n=2,\ n=3,\ n=4,\ n=5\ \text{to the equation}:\\\\n=1\to a_1=72\left(\dfrac{1}{3}\right)^{1-1}=72\left(\dfrac{1}{3}\right)^0=72(1)=72\\\\n=2\to a_2=72\left(\dfrac{1}{3}\right)^{2-1}=72\left(\dfrac{1}{3}\right)^1=72\left(\dfrac{1}{3}\right)=\dfrac{72}{3}=24\\\\n=3\to a_3=72\left(\dfrac{1}{3}\right)^{3-1}=72\left(\dfrac{1}{3}\right)^2=72\left(\dfrac{1}{9}\right)=\dfrac{72}{9}=8\\\\n=4\to a_4=72\left(\dfrac{1}{3}\right)^{4-1}=72\left(\dfrac{1}{3}\right)^3=72\left(\dfrac{1}{27}\right)=\dfrac{72}{27}=\dfrac{8}{3}\\\\n=5\to a_5=72\left(\dfrac{1}{3}\right)^{5-1}=72\left(\dfrac{1}{3}\right)^4=72\left(\dfrac{1}{81}\right)=\dfrac{72}{81}=\dfrac{8}{9}\\\\Answer:\ \boxed{72,\ 24,\ 8,\ \dfrac{8}{3},\ \dfrac{8}{9}}

8 0
3 years ago
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