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3 years ago

What ordered pair is a solution of the equation ? 2x + 4y = 6x - y

Mathematics

1 answer:

marusya05 [52]3 years ago

4 0

Use these pictures to figure it out also download Photomath

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Paul travel to the lake and back. The trip took 3 hours and the trip back took 4 hours He averaged 10 mph faster on the trip the

ollegr [7]

Answer:

40 mph

Step-by-step explanation:

We assume "outbound" refers to the trip <em>to the lake</em>. The ratio of speeds is inversely proportional to the ratio of times, so ...

outbound speed : inbound speed = 4 : 3

These differ by one ratio unit, so that one ratio unit corresponds to the speed difference of 10 mph. Then the 4 ratio units of outbound speed will correspond to ...

4×10 mph = 40 mph

Paul's average speed on the outbound trip was 40 mph.

___

The distance to the lake was 120 mi.

4 0

3 years ago

Read 2 more answers

Why 2 is not a solution of x>5

NikAS [45]

Answer:

X has to be bigger than 5 and 2 is lower than 5.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Kylie draws a triangle on a coordinate grid. She then performs the following

zmey [24]

B PLEASE GIVE BRAINLIEST

7 0

3 years ago

Solve the system. And explain please cause I have 4 more

Nady [450]

Answer:

Wong was conducting a field study in the Amazon rainforest when he

discovered a prokaryotic organism that has a cell wall and circular

chromosomes. Which of the

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2 years ago

Let v1=(3,5) and v2=(-4,7).

Nat2105 [25]

PART A

The given vectors are,

$v_1 = \: < \: 3 , \: 5 \: >$

$v_2 = \: < \: - 4 , \: 7 \: >$

The magnitude of the vector

$v= \: < \: x , \: y \: > \:$

is given by:

$|v| = \sqrt{ {x}^{2} + {y}^{2} }$

This implies that,

$|v_1| = \sqrt{ {3}^{2} + {5}^{2} }$

$|v_1| = \sqrt{9 + 25}$

$|v_1| = \sqrt{34}$

$|v_2| = \sqrt{ {( - 4)}^{2} + {7}^{2} }$

$|v_2| = \sqrt{ 16+ 49}$

$|v_2| = \sqrt{65}$

PART B

To find the unit vector in the direction of a given vector, we divide by the magnitude of that vector.

$^{ - } _{v_1} = \: < \: \frac{3}{ \sqrt{34} } , \: \frac{5}{ \sqrt{34} } \: >$

Rationalize the denominator.

$^{ - } _{v_1} = \: < \: \frac{3\sqrt{34}}{ 34 } , \: \frac{5\sqrt{34}}{ 34 } \: >$

Also,

$^{ - } _{v_2} = \: < \: \frac{ - 4}{ \sqrt{65} } , \: \frac{7}{ \sqrt{65} } \: >$

$^{ - } _{v_2} = \: < \: \frac{ - 4\sqrt{65}}{ 65 } , \: \frac{7\sqrt{65}}{ 65 } \: >$

PART C

The sketch of the given vectors as well as their unit vectors are shown in the attachment.

3 0

3 years ago