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3 years ago

Please answer right now there can be more than one option answer in two minutes

Mathematics

1 answer:

Iteru [2.4K]3 years ago

3 0

Answer:

i believe the correct answer would be the all of them

Step-by-step explanation:

they all have shared points

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Please Help me this is the first part of 4 in this question. I will be posting the other 3 shortly. I need the answer. This is v

Ket [755]

Answer:

y = 1/6x + 5

Step-by-step explanation:

-x + 6y = 30

6y = x + 30

y = 1/6x + 5

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2 years ago

The length of a rectangle is 24 units. Can the perimeter x of the rectangle be 60 units when its width y is 11 units? (1 point)

zlopas [31]

The first choice is correct. No, the rectangle cannot have x=60 and y=11 because x=48+2y.

Perimeter is x=2w+2y. Therefore x=2(24)+2(11) which equals 70 not 60.

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3 years ago

A basketball player shoots the ball with an initial velocity of 17.1 ft/sec at an angle of 45.6 degrees with the horizontal.to t

jeka94

The horizontal component is:

17.1cos45.6≈11.96 so

vx=12.0 ft/sec (to nearest tenth ft/s)

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3 years ago

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Anyone help me wiv this question please

Jlenok [28]

Answer:

A: 8x-12

B: 5(n+3)

Step-by-step explanation:

A) Distribute 4 to the numbers inside the parenthesis.

4 x 2x= 8x

4 x -3= -12

8x-12

B) 5 is divisible to 15, so you can divide it with both numbers.

5/5=1

15/5=5

5(n+3)

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3 years ago

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From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

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3 years ago

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