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4 years ago

Which ordered pair is a solution of the equation

Mathematics

1 answer:

alekssr [168]4 years ago

8 0

Answer:

OPTION D: NEITHER

Step-by-step explanation:

The given equation is: 7x - 2y = - 5

To find a solution to this, we substitute the options and compare LHS and RHS.

OPTION A: (1, 5)

LHS = 7(1) - 2(5) = 7 - 10 = -3

RHS = - 5

LHS $\ne$ RHS.

So, this option is eliminated.

OPTION B: (-1, 1)

LHS = 7(-1) - 2(1) = -7 - 2 = - 9

RHS = - 5

Again, LHS $\ne$ RHS.

So, this Option is eliminated as well.

OPTION C: It says both A and B. Clearly, this is eliminated as well.

Therefore, the answer is: OPTION D: NEITHER.

NOTE: This is a two variable equation. So, we need a minimum of two equations to determine the solution. Since, only one equation is given here, we use the help of options.

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3 years ago

A book is in the shape of a right rectangular prism. It measures 8 1/2 inches tall, 7 inches wide, and 2 1/2 inches thick. What

Alenkasestr [34]

Hello!

The formula for volume of a rectangular prism is:

V = lwh

Let's convert the dimensions to decimals, and then use the formula to solve.

8 1/2 = 8.5 as a decimal.

7 = 7 as a decimal.

2 1/2 = 2.5 as a decimal.

Multiply:

V = 8.5 × 7 × 2.5

V = 59.5 × 2.5

V = 148.75

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ANSWER:

The volume of the book is 148 3/4 inches cubed.

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3 years ago

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Classify the model as exponential growth or exponential decay. Identify the growth or decay factor AND the percent of increase o

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The growth factor is 1.05.

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3 years ago

NEED GEOMETRY HELP ASAP PLEASE (12 POINTS)

sergij07 [2.7K]

Answer:

2 times the square root of 10

Step-by-step explanation:

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3 years ago

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CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

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3 years ago