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3 years ago

Math stuff can someone help a girl out :D one to one function stuff

Mathematics

1 answer:

Anna11 [10]3 years ago

5 0

Answer:

It is not one to one... Because if you draw horizontal lines, it interests the graph at two points

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(3v^5-34v^4+79v^3+5v^2+21v+19)\(v-8)

malfutka [58]

Answer:

3 big guys

Step-by-step explanation:

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2 years ago

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SOMEONE PLS I NEED HELP ASAP!!! I WILL GIVE BRAINLIEST AND POINTS!!!

Olin [163]

Answer:

The resultant speed is 294.5 mi/h in a direction N79.7°E

Step-by-step explanation:

Let

East and North as the positive x and y-axis, respectively.

West and South as the negative x- and y-axis, respectively.

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3 years ago

Find the great common factor of 33, 63, and 90.

Alisiya [41]

Answer:

GCF is 3

Step-by-step explanation:

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3 years ago

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

5 0

3 years ago

Part A

nydimaria [60]

Answer:

1) D

2) 10

Step-by-step explanation:

sin(30) = opposite/hypotenuse

sin(30) = 5/x

½ = 5/x

x = 5/0.5

x = 10

5 0

3 years ago