All categories

2 years ago

When the equation 16/x = 4 is simplified, the value of x is??

Mathematics

2 answers:

Darina [25.2K]2 years ago

5 0

The value of x is 4.

astraxan [27]2 years ago

5 0

Answer: x=4

Step-by-step explanation:

multiply each side by the denominator to cancel out the fraction so the new equation is 16=4x then divide by 4 to eliminate the x and 16 divided by 4 is 4 which is the anwser.

You might be interested in

Write an equivalent expression for each of the following. Express your answers in the form a + bi or a - bi

Lana71 [14]

Answer:

44 +14i

Step-by-step explanation:

(5 + i)(9 + i)

FOIL

first 5*9 = 45

inner: 9i

outer :5i

last: i*i = i^2 = -1

Add togheter

45 + 9i+5i -1

Combine like terms

44 +14i

6 0

3 years ago

From 1000 students half being boys and half being girls 100 names were pulled out of a hat to complete a survey for school what

Mama L [17]

Random sampling. its where you pick or make a smaple at random

6 0

3 years ago

Read 2 more answers

What is the relationship between the values tand s plotted on the number line

vivado [14]

A. T=s
Im pretty sure

4 0

3 years ago

Read 2 more answers

I don’t know the answer I just need some points to ask more questions !!! have a good day !!

Dahasolnce [82]

Answer:

what ??

Step-by-step explanation:

5 0

3 years ago

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

tangare [24]

Answer:

$AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm$

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

$AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm$

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

$\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD$

Consider right triangle CAH.By the Pythagorean theorem,

$CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8$

The length cannot be negative, so CD=8 cm and

$CA=\sqrt{3}CD=8\sqrt{3}\ cm$

In right triangle ABC, angle B = 90° - 60° = 30°, leg AC is opposite to 30°, and the hypotenuse AB is twice the leg AC. Hence,

$AB=2CA=16\sqrt{3}\ cm$

By the Pythagorean theorem,

$BC^2=AB^2-AC^2\\ \\BC^2=(16\sqrt{3})^2-(8\sqrt{3})^2=256\cdot 3-64\cdot 3=576\\ \\BC=24\ cm$

3 0

3 years ago