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3 years ago

What are the solutions to the equation?

Mathematics

2 answers:

Anika [276]3 years ago

6 0

The answer to your question is the last one through the method of competing the square

AleksandrR [38]3 years ago

6 0

Answer:

x = −10 and x = 4

Step-by-step explanation:

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A motorboat, that has speed of 10 km/hour in still water, left a pier traveling against the current of the river. Forty-five min

Nutka1998 [239]

Answer:

2 km/h

Step-by-step explanation:

distance = speed × time
time = distance/speed

Let c represent the speed of the current of the river in km/h. Then the speed of the boat upstream is (10-c). In 3/4 hour, the distance traveled upstream is ...

... distance upstream = (3/4)·(10 -c)

The time taken to travel the same distance downstream is given as 3 hours. The speed in that direction is c, so we have ...

... 3 = distance upstream/c = (3/4)(10 -c)/c

Multiplying this equation by 4c, we get ...

... 12c = 3(10 -c)

... 15c = 30 . . . . . . . . add 3c

... c = 2 . . . . . . . . . . . divide by 15

The speed of the river current is 2 km/h.

7 0

4 years ago

NEED HELP ASAP! WILL MARK BRAINLIEST!

kenny6666 [7]

Answer:

y=2x-20

Step-by-step explanation:

For perpendicular lines, find the opposite reciprocal of the original slope. The opposite reciprocal of the slope of this line (-1/2) is positive 2

7 0

3 years ago

14, 23, 31, 29, 33 what is the standard deviation for this set of population data?

Readme [11.4K]

31 would be the standard

5 0

3 years ago

7h + 3 = 7(2=h) -11 <br> this is my question what is the answer???

NISA [10]

Answer:

7h+3=-11?

then h=-11-3 divided by 7

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your

Mashutka [201]

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

$D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Distance between two points P(-2,5) and Q(-1,1)

$PR = \sqrt{(-1+2)^{2}+(1-5)^{2} }$

$PR = \sqrt{17}$

Distance between two points Q(-1,1)and R(7,3)

$QR = \sqrt{(7+1)^{2} +(3-1)^{2} }$

$QR =\sqrt{68}$

Distance between two points R(7,3) and P(-2,5)

$RP =\sqrt{(-2-7)^{2} + (5-3)^{2} }$

$RP=\sqrt{85}$

now show that ∆PQR is a right triangle

$RP^{2} = PQ^{2} +QR^{2}$

Putting the value given above

$(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}$

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved

7 0

4 years ago