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1 year ago

How does 2 relate to 3

Mathematics

1 answer:

erik [133]1 year ago

4 0

Answer:

The number that was added to the previous number is doubled.

For example:

2+1 = 3

2x1= 2

3+2= 5

2x2=4

5+4=9

4x2=8

8+9=17

I hope this helps:)

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Carlos has 4.5 pounds of flour she needs three out of 4 pound of flour to make pizza what is the greatest number of pizzas Carlo

iogann1982 [59]

Answer:

6 pizzas

Step-by-step explanation:

divide 4.5 by 3/4

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2 years ago

Sam is a forklift operator at a Los Angeles warehouse. A 500 kg crate arrived from Taiwan. His forklift can load 2,500 lb. He wa

Wittaler [7]

1 kg = 2.2 pounds ( rounded to the nearest tenth)

500 kg x 2.2 = 1,100 pounds

1100 pounds is less than 2,500 so the forklift can lift it.

8 0

2 years ago

Read 2 more answers

PLEASE HELP! GIVING BRAINLIST!!

erica [24]

Answer:

When x = 1, y = 10.

Step-by-step explanation:

Well, the graph is exponential. But in terms of the question, you can say if there is 1 year increase, there are 10 snakes.

4 0

1 year ago

Is the open sentence 3z=2z+5 true or false when z=5?<br><br> A. True<br> B. False

inna [77]

Hey there!

If z = 5

we replace "z" with "5"

So, let's test out the problem to see if it is true or false

Problem becomes: $3(5) = 2(5) + 5$

Let's work this out

$3(5) = 15$

Problem becomes: $15 = 2(5) + 5$

$2(5) = 10$

Problem becomes: $15 = 10 + 5$

$10 + 5 = 15$

So, therefore it looks like this statement/ sentence is $True$ !

~ $MeIsKaitlyn:)$

3 0

2 years ago

I need you to answer with a, b, c, d

solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

$\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

For the given function:

$f(x)=2x^2-10x-3$ $x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}$ $x=\frac{10\pm\sqrt[]{100+24}}{4}$ $\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}$ $\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}$ $\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Then, the zeros of the given quadratic function are:

$\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\ \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Answer: Third option

8 0

10 months ago