2x-8+3x(greater than or equal to)5x-4

masha68 [24]

3x² - 33x - 180 = 0
3(x² - 11x - 60) = 0
3(x² - 15x + 4x - 60) = 0
3(x + 4)(x - 15) = 0
3(x - 4) = 0 or x - 15 = 0
3x - 12 = 0 + 15 + 15
 + 12 + 12 x = 15
 3x = 12
 3 4
 x = 3

The answer is equal to H.15 and E. 4.

8 0

3 years ago

Can someone help me with this. Will Mark brainliest. Need answer and explanations/work. Thank you.

hoa [83]

Answer:

cosine of angle a = 8/17

Step-by-step explanation:

Hello there!

Remember these are the trigonometric ratios

SOC CAH TOA

Sine = Opposite over Hypotenuse (SOH)

Cosine = Adjacent over Hypotenuse (CAH)

Tangent = Opposite over Adjacent (TOA)

And we are asked to find the Cosine of angle A

Remember cosine is adjacent over hypotenuse

Hypotenuse - The longest side

Adjacent - the side that's not the hypotenuse nor the opposite

The adjacent side length of angle A is equal to 8 and the hypotenuse is equal to 17

so the cosine of angle a = 8/17

3 0

3 years ago

if you stop at random in front of an animal exhibit,What is the probability that you will be looking at monkeys? I put the table

Mkey [24]

Answer:

1/10

step-by-step explanation:

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| animals | percent of animals |

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| goats | 16% |

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| camels | 4% |

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| giraffes | 6% |

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| llamas | 6% |

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| deer | 1% |

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| monkeys | 10% |

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| birds | 27% |

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| reptiles | 14% |

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| bears | 16% |

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( sorry if the table above isnt 100% right or straight ^ )

first you add up all the percents of how much animals are in this exhibit

16 + 4 + 6 + 6 + 1 + 10 + 27 + 14 + 16

= 100 %

then since we are looking for the probability of stopping in front of the monkeys, we take that 100 we got earlier and get the 10% of monkeys from our table

10/100

but since there isnt a 10/100 on the answers we look for something that equals to 10/100 and 1/10 does since you can just times both the 1 and 10 to equal the 10 and 100 you were looking for

1/10

making 1/10 your answer!

hope this helps! and if you need anymore help on this math question just ask me :)

4 0

2 years ago

Calculate the number of kilograms of mercury required to fill a pool that is 20.0 m x 50.0 m x 10.0 m. The density of mercury is

malfutka [58]

Answer:

$m=135000000 Kg$

Step-by-step explanation:

To find the number of kilograms of mercury we need to find how to relate density, mass and, volume. For this we shall recall the density formula:

$\rho=\frac{m}{V}$

where $\rho$ is the density, $m$ is the mass and, $V$ is the volume.

We have the density and want to compute the mass so now we want to know the volume of the pool.

The volume of a rectangular pool is given by the fomula:

$V= (length)(heigh)(width)$ .

So for our pool

$V= (20)(50)(10)$ .

$V= 10000m^{3}$ .

Our density is in $g/mL$ , so the last thing we need to do before computing the mass is to express the density in $Kg/m^{3}$ (this is because we want our mass in $Kg$ and, we have our volume in $m^{3}$ ).