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3 years ago

Which equation represents the line that is perpendicular to and passes through (-4,-3)?

Mathematics

1 answer:

nignag [31]3 years ago

8 0

Answer:

I think It's x=-4

Step-by-step explanation:

Hope this helps :))

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when rolling a number cube numbered 1-6 what is the probability you will roll a 3,4,or 5 three times in a row

vodka [1.7K]

The probability is 50% or 1/2

6 0

3 years ago

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Brahmagupta’s solution to a quadratic equation of the form ax2 + bx = c involved only one solution. Which solution would he have

zlopas [31]

Answer:

0.897

Step-by-step explanation:

Brahmagupta formula for quadratic equation $ax^2+bx=c$ is

$x=\dfrac{\sqrt{4ac+b^2}-b}{2a}$

It involved only one solution.

The given equation is

$3x^2+4x=6$

Here, a=3, b=4 and c=6. Put these values in the above formula.

$x=\dfrac{\sqrt{4(3)(6)+(4)^2}-4}{2(3)}$

$x=\dfrac{\sqrt{72+16}-4}{6}$

$x=\dfrac{\sqrt{88}-4}{6}$

$x\approx \dfrac{5.38}{6}$

$x\approx 0.897$

Therefore, the required solution is 0.897.

7 0

3 years ago

Your soccer team has won 9 games out of 12 games. You lost 3 out of the 12 games. What is the relative frequency of winning a ga

LiRa [457]

Answer:

75%

Step-by-step explanation:

9 ÷ 12 = 0.75

0.75 = 75%

6 0

3 years ago

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A news report says that 9%9% of middle school students are on the track team. There are 600600 students in your middle school. H

andrew11 [14]

Assuming you meant 600 students, the answer would be 54 on the track team.

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3 years ago

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At noon a tank contains 6 cm of water after several hours it contains 3 cm of water what is the percent inecrease of water in th

Hitman42 [59]

Question:

At noon a tank contains 6 cm of water after several hours it contains 3 cm of water what is the percent decrease of water in the tank

Answer:

The percent Decrease of water in the tank is 50%

Step-by-step explanation:

Given:

Initial amount of water in the tank at the noon = 6 cm

amount of water remained in water after several hours = 3 cm

To Find:

The percent decrease =?

Solution:

The percent decrease in the tank

=> $\frac{\text {water present after some hours }}{\text {initial water contained in the tank }} \times 100$

Substituting the values

=> $\frac{3}{6} \times 100$

=> $\frac{1}{2} \times 100$

=> $\frac{100}{2}$

=> 50

5 0

3 years ago