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2 years ago

13

PLEASE help me with this I really need help...

1 answer:

svetoff [14.1K]2 years ago

3 0

<h3>Find out the initial distance covered by the pumpkin at the instance of release. </h3>

<h3>For that, </h3><h3>Put x=0 in each option and compare it with the value that you'll get after putting x=0. in the given equation. </h3>

<h3>The option that gives more value than the value of the equation will be the answer. </h3>

<h3><em>Followi</em><em>ng</em><em> </em><em>the </em><em>abo</em><em>ve</em><em> </em><em>rule</em><em> </em><em>.</em><em>.</em><em>.</em><em> </em><em>I </em><em>think</em><em> </em><em>Option</em><em> </em></h3><h3><em>A</em><em> </em><em>shou</em><em>ld</em><em> </em><em>be</em><em> </em><em>corr</em><em>ect</em><em>.</em><em> </em></h3>

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Write 17% as a decimal

netineya [11]

0.17

yuh this is correct lol

5 0

2 years ago

Heidi has 6 pounds of tomatoes from her garden. She used 3/4 of all the tomatoes to make sauce and gave 2/3 of the rest of the t

iren2701 [21]

Answer:

She gives 16 ounces to her neighbor

Step-by-step explanation:

How much did you use to make sauce

6 * 3/4 = 18/4 =9/2 = 4.5 lbs

That leaves 6-4.5 = 1.5 lbs

She give 2/3 of this to her neighbor

1.5 * 2/3

Changing the decimal to a fraction

3/2 * 2/3 = 1

She gives 1 pound to her neighbor, but we want it in ounces

16 ounces equals 1 pound

She gives 16 ounces to her neighbor

6 0

3 years ago

Sequences HW<br><br> Please helpp<br><br> Thank you

Alexus [3.1K]

Answer:

0.30

Step-by-step explanation:

Each number was getting divided by 3 each time

6 0

2 years ago

Read 2 more answers

Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)

masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

$y=ax^2+bx+c$ ...(i)

It is passes through the point (0,11). So, substitute $x=0,y=11$ in (i).

$11=a(0)^2+b(0)+c$

$11=c$

Putting $c=11$ in (i), we get

$y=ax^2+bx+11$ ...(ii)

The quadratic function passes through the point (5,31). So, substitute $x=5,y=31$ in (ii).

$31=a(5)^2+b(5)+11$

$31-11=a(25)+5b$

$20=25a+5b$

Divide both sides by 5.

$4=5a+b$ ...(iii)

The quadratic function passes through the point (3,11). So, substitute $x=3,y=11$ in (ii).

$11=a(3)^2+b(3)+11$

$11-11=a(9)+3b$

$0=9a+3b$

Divide both sides by 3.

$0=3a+b$ ...(iv)

Subtracting (iv) from (iii), we get

$4-0=5a+b-3a-b$

$4=2a$

$\dfrac{4}{2}=a$

$2=a$

Putting $a=2$ in (iv), we get

$0=3(2)+b$

$0=6+b$

$-6=b$

Putting $a=2,b=-6$ in (ii), we get

$y=(2)x^2+(-6)x+11$

$y=2x^2-6x+11$

Therefore, the required quadratic equation is $y=2x^2-6x+11$ .

7 0

2 years ago

Can you help me with this math problem

Marat540 [252]

Answer:

It might be 70 because 5m has the volume of 50 m so maybe 7m in volume is 70?

3 0

2 years ago