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3 years ago

Which of the following rules could represent the function shown in the table?

Mathematics

1 answer:

MariettaO [177]3 years ago

6 0

Answer:

f(x) = 2x+1

Step-by-step explanation:

Equation in slope intercept form y = mx + b where m = slope b = y-intercept

To find slope ,

m = (y2-y1)/(x2 - x1)

m = (3 - 1) / (1 - 0)

m = 32/1

m = 2

y-intercept b = 1

In this case y = 2x + 1 or f(x) = 2x+1

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Yamri went to the store to buy gardening supplies. A bag of dirt was $3.99 and tulips cost 75 cents per bulb. He bought one bag

seraphim [82]

3.99 + 0.75b = 12.24
0.75b = 12.24 - 3.99
0.75b = 8.25
b = 8.25 / 0.75
b = 11.....there were 11 bulbs

8 0

3 years ago

Read 2 more answers

Solve the following system:

Angelina_Jolie [31]

You can rewrite the second equation because you know what y equals, so you can write it as 2x + x + 3 = 9
3x + 3 = 9
- 3
3x = 6
÷ 3
x = 2

And now you use the equation of what y equals and substitute in the value of x, so y = 2 + 3
y = 5

So your final answer is A. (2, 5). I hope this helps!

7 0

4 years ago

Given BE= 2x+6 and ED= 5x-12 in a parallelogram ABCD, find BD

Brrunno [24]

1.BE = 2x + 6
ED = 5x - 12
2. To get the entire side of BD, we must add both half's which equals to the entire length.
3. 2x + 6 + 5x - 12
4.Add like terms.
2x + 5x = 7x
-12 + 6 = -6
5. So, we have 7x - 6

=7x - 6

3 0

4 years ago

You're standing 80 feet from the base of a building . You estimate that angle of elevation from your feet to the top is about 70

pogonyaev

Probably about (at least) 120 feet.

6 0

3 years ago

Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile

mario62 [17]

Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships is 135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC = 402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40' [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

$\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Taking last two ratio,

$\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Putting the value of BC , AC ,∠A,∠B

$\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}$

$\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}$

≈734.106 miles

Therefore they are 734.106 miles apart.

3 0

3 years ago