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3 years ago

What is the slipe of (-1,3) and (1,3) ?

Mathematics

1 answer:

Minchanka [31]3 years ago

7 0

Answer:

0 theres no slope lol cause

3-3= 0

1--1=2

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Evaluate the expression 2x +y for x=7and y=6

Andre45 [30]

Answer: 20

Step-by-step explanation:

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3 years ago

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Point G is the centroid of the right △ABC with m∠C=90° and m∠B=30°. Find AG if CG=4 ft.

o-na [289]

Answer: $\text{Length of AG=}\frac{2\sqrt{63}}{3}$

Explanation:

Please follow the diagram in attachment.

As we know median from vertex C to hypotenuse is CM

$\therefore CM=\frac{1}{2}AB$

We are given length of CG=4

Median divide by centroid 2:1

CG:GM=2:1

Where, CG=4

$\therefore GM=2$ ft

Length of CM=4+2= 6 ft

$\therefore CM=\frac{1}{2}AB\Rightarrow AB=12$

In $\triangle ABC, \angle C=90^0$

Using trigonometry ratio identities

$AC=AB\sin 30^0\Rightarrow AC=6$ ft

$BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3}$ ft

$CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3}$ ft

In $\triangle CAN, \angle C=90^0$

Using pythagoreous theorem

$AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}$

Length of AG=2/3 AN

$\text{Length of AG=}\frac{2\sqrt{63}}{3}$ ft

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3 years ago

I WILL GIVE BRAINLIEST 6TH GRADE MATH

yaroslaw [1]

Answer:

Step-by-step explanation: If you count a the answer will be 48

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2 years ago

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Given the infirmation in the diagram, which theorem best justifies why lines j and k must be parallel

Karo-lina-s [1.5K]

Answer:

Alternate Exterior Angles theorem

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3 years ago

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Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 2,250,000 and a mean

Svet_ta [14]

Answer: 0.1357

Step-by-step explanation:

Given : Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of $\sigma^2=2,250,000$ and a mean life span of $\mu=13,000$ hours.

Here , $\sigma=\sqrt{2250000}=1500$

Let x represents the life span of a monitor.

Then , the probability that the life span of the monitor will be more than 14,650 hours will be :-

$P(x>14650)=P(\dfrac{x-\mu}{\sigma}>\dfrac{14650-13000}{1500})\\\\=P(z>1.1)=1-P(z\leq1.1)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\=1-0.8643339=0.1356661\approx0.1357$

Hence, the probability that the life span of the monitor will be more than 14,650 hours = 0.1357

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3 years ago