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3 years ago

Which coordinates best represents the solution to the following pair of equations?

1 answer:

7 0

Answer is B. For y = -2x + 13 If we put x=4 y = -2(4) + 13 y = -8 + 13 y = 5 So it becomes (4,5) For y = 2x - 3 If we put x=4 y = 2(4) - 3 y = 8 - 3 y = 5 And it becomes (4,5) So (4,5) satisfies both the equations.

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Answer a, b and c. See image below

dlinn [17]

Answer:

a) 3/5 < 4/5

b) In general if two fractions have the same denominator, then whichever fraction has the numerator closer to its denominator will be the largest fraction.

c) $\frac{7}{10} > \frac{9}{15}$ or $\frac{7}{10}$

Step-by-step explanation:

a) 3/5 < 4/5

Flip the sign and the placement of the fraction so 3/5 is less then 4/5.

b) In general if two fractions have the same denominator, then whichever fraction has the numerator closer to its denominator will be the largest fraction.

c) We need to change the denominators to a common denominator to compare the size of the two fractions:

$\frac{7}{10}$ × $\frac{3}{3}$ = $\frac{21}{30}$

$\frac{9}{15}$ × $\frac{2}{2}$ = $\frac{18}{30}$

The common denominators of the two fractions is 30. Comparing the two fractions:

$\frac{21}{30} >\frac{18}{30}$ or $\frac{18}{30}$

so we get: $\frac{7}{10} > \frac{9}{15}$ or $\frac{7}{10}$

7 0

3 years ago

Please help meeeee xoxx

docker41 [41]

Answer:

b)4 11/10

Step-by-step explanation:

4 0

2 years ago

Solve: 3x - 2x + 4 = 5x - 4x - 8 A) 3 B) 6 C) -3 Eliminate D) no solution

SSSSS [86.1K]

There is no solution

5 0

3 years ago

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$

7 0

3 years ago

Read 2 more answers

How can you compare ratios to solve a problem?

Sindrei [870]

“To use proportions to solve ratio word problems, we need to follow these steps:
Identify the known ratio and the unknown ratio.
Set up the proportion.
Cross-multiply and solve.
Check the answer by plugging the result into the unknown ratio.”
I hope this helps.

6 0

3 years ago

Read 2 more answers