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3 years ago

14

The plane that contains points C and T can also be named plane...

1 answer:

Mnenie [13.5K]3 years ago

6 0

Answer:

A) CUB

Step-by-step explanation:

Of the suggested planes, only CUB contains both points C and T.

___

<em>Comments on the other answer choices</em>

BED contains point T, but not C

ACE contains point C, but not T

ABE contains neither C nor T

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Can anyone help me on this and pls explain this

7nadin3 [17]

I can’t really see it well

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3 years ago

Help!!! What’s the approximate sector area

-BARSIC- [3]

Answer:

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Step-by-step explanation:

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3 years ago

Y= x^2 -5 solve for x

dexar [7]

$Y= x^2 -5$

We need to solve for x, we need to get x alone

$Y= x^2 -5$

Lets start by removing -5

Add 5 on both sides

$y + 5= x^2 -5 + 5$

$y + 5= x^2$

Now to isolate x , we need to remove the square from x

To remove square , take square root on both sides

$+-\sqrt{y+5} = \sqrt{x^2}$

square and square root will get cancelled

$+-\sqrt{y+5} = x$

So $x = +\sqrt{y+5}$ and $x = -\sqrt{y+5}$

7 0

3 years ago

A piece of cardboard has the dimensions (x + 15) inches by (x) inches with the area of 60 in 2 . Write the quadratic equation th

pochemuha

Answer:

$x^2 + 15x - 60 = 0$

The actual dimension is 18.28 by 3.28

Step-by-step explanation:

Given

Dimension:

$(x + 15)\ by\ x$

$Area = 60in^2$

Required

Determine the quadratic equation and get the possible values of x

Solving (a): Quadratic Equation.

The cardboard is rectangular in shape.

Hence, Area is calculated as thus:

$Area = Length * Width$

$60= (x + 15) * x$

Open Bracket

$60= x^2 + 15x$

Subtract 60 from both sides

$x^2 + 15x - 60 = 0$

<em>Hence, the above represents the quadratic equation</em>

Solving (b): The actual dimension

First, we need to solve for x

This can be solved using quadratic formula:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where

$a = 1$

$b = 15$

$c = -60$

So:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-15 \± \sqrt{15^2 - 4*1*-60}}{2*1}$

$x = \frac{-15 \± \sqrt{225 + 240}}{2}$

$x = \frac{-15 \± \sqrt{465}}{2}$

$x = \frac{-15 \± \21.56}{2}$

Split:

$x = \frac{-15 + 21.56}{2}$ or $x = \frac{-15 - 21.56}{2}$

$x = \frac{6.56}{2}$ or $x = \frac{-36.36}{2}$

$x = 3.28$ or $x = -18.18$

But length can't be negative;

So:

$x = 3.28$

The actual dimensions: $(x + 15)\ by\ x$ is

$Length =3.28 +15$

$Length =18.28$

$Width = x$

$Width =3.28$

<em>The actual dimension is 18.28 by 3.28</em>

3 0

3 years ago

A factory produces car tires. The number of miles each tire lasts before it completely wares out follows a normal distribution w

qwelly [4]

Answer: 0.0668073

Step-by-step explanation:

Given : The number of miles each tire lasts before it completely wares out follows a normal distribution with mean μ = 50,000 miles and standard deviation σ = 8,000 miles.

Let x be the random variable that represents the number of miles each tire lasts.

z-score : $\dfrac{x-\mu}{\sigma}$

For x= 62,000

$z=\dfrac{62000-50000}{8000}=1.5$

By using the standard z-value table , the probability for a randomly selected tire to last for at least 62,000 miles will be :_

$P(x\geq62000)=P(z\geq1.5)=1-P(z$

Hence, the probability for a randomly selected tire to last for at least 62,000 miles = 0.0668073

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3 years ago