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3 years ago

Which of the following systems of linear equations has an infinite number of solutions?

Mathematics

1 answer:

arlik [135]3 years ago

8 0

Answer:

{ y = 3x + 3

{ 3x - y = 3

(Just did the assignment 3/23/21)

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Write the standard equation for the circle

saw5 [17]

Answer:

The standard form of a circle's equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius.

Step-by-step explanation:

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2 years ago

Please help If 25% of some number is 40, then what is 75% of that number?

Nikitich [7]

Explanation : % is the symbol for percentage. It means out of 100. 75% means 75 out of 100 which is represented as 75/100.

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3 years ago

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$2.95 notebooks; 5% tax

dimaraw [331]

Turn 5% into a decimal which is 0.05, then multiply by 2.95 = 0.1475
round it off, which is 0.15
so $0.15 is the tax and if ur looking for the total price then add $2.95 to $0.15

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3 years ago

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Please please help me. i’ll literally send you money. i need to pass please help

mihalych1998 [28]

<h3>Answer: Approximately 6.4 units</h3>

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

Explanation:

The origin is the point (0,0)

Use the distance formula to find the distance from (0, 0) to (4, -5)

Let

$(x_1,y_1) = (0,0)\\\\(x_2,y_2) = (4,-5)\\\\$

be our two points. Plug those values into the distance formula below and use a calculator to compute

$d = \sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0-(-5)\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0+5\right)^2}\\\\d = \sqrt{\left(-4\right)^2+\left(5\right)^2}\\\\d = \sqrt{16+25}\\\\d = \sqrt{41} \ \text{ exact distance}\\\\d \approx 6.40312 \ \text{ approximate distance}\\\\d \approx 6.4\\\\$

The distance between the two points (0,0) and (4,-5) is approximately 6.4 units.

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4 years ago

A ^-4 / a ^-6 please simplify and show your work if possible <:D

kipiarov [429]

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{a^{-4}}{a^{-6}}\implies a^{-4}\cdot a^6\implies a^{-4+6}\implies a^2$

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3 years ago